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User:Cornince

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Alternative account: User:Beneficii

Basic definition of a sum

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∑ i = m n f ( i ) = f ( m ) + f ( m + 1 ) + f ( m + 2 ) + . . . + f ( n − 2 ) + f ( n − 1 ) + f ( n ) {\displaystyle \sum _{i=m}^{n}{\mathit {f}}(i)={\mathit {f}}(m)+{\mathit {f}}(m+1)+{\mathit {f}}(m+2)+,円...,円+{\mathit {f}}(n-2)+{\mathit {f}}(n-1)+{\mathit {f}}(n),円} {\displaystyle \sum _{i=m}^{n}{\mathit {f}}(i)={\mathit {f}}(m)+{\mathit {f}}(m+1)+{\mathit {f}}(m+2)+,円...,円+{\mathit {f}}(n-2)+{\mathit {f}}(n-1)+{\mathit {f}}(n),円}


Recursive summation

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Where p ≥ 0 {\displaystyle p\geq 0} {\displaystyle p\geq 0}:

∑ i = m i p p   f ( i ) = ∑ i p − 1 = m i p ∑ i p − 2 = m i p − 1 ∑ i p − 3 = m i p − 2 . . . ∑ i 2 = m i 3 ∑ i 1 = m i 2 ∑ i 0 = m i 1 f ( i 0 ) {\displaystyle \sum _{i=m}^{i_{p}}\!{}^{p}\ f(i)=\sum _{i_{p-1}=m}^{i_{p}},円\sum _{i_{p-2}=m}^{i_{p-1}},円\sum _{i_{p-3}=m}^{i_{p-2}}...\sum _{i_{2}=m}^{i_{3}},円\sum _{i_{1}=m}^{i_{2}},円\sum _{i_{0}=m}^{i_{1}}f(i_{0}),円} {\displaystyle \sum _{i=m}^{i_{p}}\!{}^{p}\ f(i)=\sum _{i_{p-1}=m}^{i_{p}},円\sum _{i_{p-2}=m}^{i_{p-1}},円\sum _{i_{p-3}=m}^{i_{p-2}}...\sum _{i_{2}=m}^{i_{3}},円\sum _{i_{1}=m}^{i_{2}},円\sum _{i_{0}=m}^{i_{1}}f(i_{0}),円}

∑ i = m i 0 0   f ( i ) = f ( i 0 ) {\displaystyle \sum _{i=m}^{i_{0}}\!{}^{0}\ {\mathit {f}}(i)={\mathit {f}}(i_{0}),円} {\displaystyle \sum _{i=m}^{i_{0}}\!{}^{0}\ {\mathit {f}}(i)={\mathit {f}}(i_{0}),円}

∑ i = m i 1 1   f ( i ) = ∑ i 0 = m i 1 f ( i 0 ) {\displaystyle \sum _{i=m}^{i_{1}}\!{}^{1}\ {\mathit {f}}(i)=\sum _{i_{0}=m}^{i_{1}}{\mathit {f}}(i_{0}),円} {\displaystyle \sum _{i=m}^{i_{1}}\!{}^{1}\ {\mathit {f}}(i)=\sum _{i_{0}=m}^{i_{1}}{\mathit {f}}(i_{0}),円}

I f   g ( n ) = ∑ i = m n − p f ( i ) ,   t h e n   f ( n ) = ∑ i = m n p g ( i ) . {\displaystyle \mathrm {If\ } {\mathit {g}}(n)=\sum _{i=m}^{n}\!{}^{-p},円{\mathit {f}}(i),\ \mathrm {then\ } {\mathit {f}}(n)=\sum _{i=m}^{n}\!{}^{p},円{\mathit {g}}(i).,円} {\displaystyle \mathrm {If\ } {\mathit {g}}(n)=\sum _{i=m}^{n}\!{}^{-p},円{\mathit {f}}(i),\ \mathrm {then\ } {\mathit {f}}(n)=\sum _{i=m}^{n}\!{}^{p},円{\mathit {g}}(i).,円}

Where p > 0 {\displaystyle p>0} {\displaystyle p>0}:

∑ i = m i p p   f ( i ) = ∑ i p − 1 = m i p [ ∑ i = m i p − 1 p − 1 f ( i ) ] {\displaystyle \sum _{i=m}^{i_{p}}\!{}^{p}\ {\mathit {f}}(i)=\sum _{i_{p-1}=m}^{i_{p}}\left[\sum _{i=m}^{i_{p-1}}\!{}^{p-1},円{\mathit {f}}(i)\right],円} {\displaystyle \sum _{i=m}^{i_{p}}\!{}^{p}\ {\mathit {f}}(i)=\sum _{i_{p-1}=m}^{i_{p}}\left[\sum _{i=m}^{i_{p-1}}\!{}^{p-1},円{\mathit {f}}(i)\right],円}

∑ i p − 1 = m i p [ ∑ i = m i p − 1 p − 1 f ( i ) ] = ∑ i = m m p − 1 f ( i ) + ∑ i = m m + 1 p − 1 f ( i ) + ∑ i = m m + 2 p − 1 f ( i ) + . . . + ∑ i = m i p − 2 p − 1 f ( i ) + ∑ i = m i p − 1 p − 1 f ( i ) + ∑ i = m i p p − 1 f ( i ) {\displaystyle \sum _{i_{p-1}=m}^{i_{p}}\left[\sum _{i=m}^{i_{p-1}}\!{}^{p-1},円{\mathit {f}}(i)\right]=\sum _{i=m}^{m}\!{}^{p-1},円{\mathit {f}}(i)+\sum _{i=m}^{m+1}\!{}^{p-1},円{\mathit {f}}(i)+\sum _{i=m}^{m+2}\!{}^{p-1},円{\mathit {f}}(i)+,円...+,円\sum _{i=m}^{i_{p}-2}\!{}^{p-1},円{\mathit {f}}(i)+\sum _{i=m}^{i_{p}-1}\!{}^{p-1},円{\mathit {f}}(i)+\sum _{i=m}^{i_{p}}\!{}^{p-1},円{\mathit {f}}(i),円} {\displaystyle \sum _{i_{p-1}=m}^{i_{p}}\left[\sum _{i=m}^{i_{p-1}}\!{}^{p-1},円{\mathit {f}}(i)\right]=\sum _{i=m}^{m}\!{}^{p-1},円{\mathit {f}}(i)+\sum _{i=m}^{m+1}\!{}^{p-1},円{\mathit {f}}(i)+\sum _{i=m}^{m+2}\!{}^{p-1},円{\mathit {f}}(i)+,円...+,円\sum _{i=m}^{i_{p}-2}\!{}^{p-1},円{\mathit {f}}(i)+\sum _{i=m}^{i_{p}-1}\!{}^{p-1},円{\mathit {f}}(i)+\sum _{i=m}^{i_{p}}\!{}^{p-1},円{\mathit {f}}(i),円}

Shifting of starting and ending indices

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∑ i = m n f ( i ) = ∑ i = m + u n + u f ( i − u ) {\displaystyle \sum _{i=m}^{n}{\mathit {f}}(i)=\sum _{i=m+u}^{n+u}{\mathit {f}}(i-u),円} {\displaystyle \sum _{i=m}^{n}{\mathit {f}}(i)=\sum _{i=m+u}^{n+u}{\mathit {f}}(i-u),円}


Proof of the equality of the shifting of indices:

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∑ i = m + u n + u f ( i − u ) = f ( ( m + u ) − u ) + f ( ( m + u + 1 ) − u ) + f ( ( m + u + 2 ) − u ) + . . . + f ( ( n + u − 2 ) − u ) + f ( ( n + u − 1 ) − u ) + f ( ( n + u ) − u ) = f ( m ) + f ( m + 1 ) + f ( m + 2 ) + . . . + f ( n − 2 ) + f ( n − 1 ) + f ( n ) = ∑ i = m n f ( i ) {\displaystyle ,円\!{\begin{array}{lcl}\sum _{i=m+u}^{n+u}{\mathit {f}}(i-u)&=&{\mathit {f}}((m+u)-u)+{\mathit {f}}((m+u+1)-u)+{\mathit {f}}((m+u+2)-u)\\&&+,円...,円+{\mathit {f}}((n+u-2)-u)+{\mathit {f}}((n+u-1)-u)+{\mathit {f}}((n+u)-u)\\\\&=&{\mathit {f}}(m)+{\mathit {f}}(m+1)+{\mathit {f}}(m+2)+,円...,円+{\mathit {f}}(n-2)+{\mathit {f}}(n-1)+{\mathit {f}}(n)\\\\&=&\sum _{i=m}^{n}{\mathit {f}}(i)\end{array}},円} {\displaystyle ,円\!{\begin{array}{lcl}\sum _{i=m+u}^{n+u}{\mathit {f}}(i-u)&=&{\mathit {f}}((m+u)-u)+{\mathit {f}}((m+u+1)-u)+{\mathit {f}}((m+u+2)-u)\\&&+,円...,円+{\mathit {f}}((n+u-2)-u)+{\mathit {f}}((n+u-1)-u)+{\mathit {f}}((n+u)-u)\\\\&=&{\mathit {f}}(m)+{\mathit {f}}(m+1)+{\mathit {f}}(m+2)+,円...,円+{\mathit {f}}(n-2)+{\mathit {f}}(n-1)+{\mathit {f}}(n)\\\\&=&\sum _{i=m}^{n}{\mathit {f}}(i)\end{array}},円}


Smaller summation notation

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∑ i = m n p   f ( i ) = p ∑ i = m n f ( i ) {\displaystyle ,円\!\sum _{i=m}^{n}\!{}^{p}\ {\mathit {f}}(i)=\textstyle \!{}_{p}\!\sum _{i=m}^{n}{\mathit {f}}(i),円} {\displaystyle ,円\!\sum _{i=m}^{n}\!{}^{p}\ {\mathit {f}}(i)=\textstyle \!{}_{p}\!\sum _{i=m}^{n}{\mathit {f}}(i),円}


Recursive geometric series

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p ∑ i = m n r i = r n + p ( r − 1 ) p − ∑ k = 0 p − 1 r m + p − ( k + 1 ) ∏ j = 1 k ( n − m + j ) k ! ( r − 1 ) p − k = r n + p ( r − 1 ) p − ∑ k = 0 p − 1 r m + p − ( k + 1 ) ( n − m + k ) ! ( n − m ) ! ( r − 1 ) p − k k ! = r n + p ( r − 1 ) p − ∑ k = 0 p − 1 ( r m + p − ( k + 1 ) ( r − 1 ) p − k ) ( n − m + k k ) {\displaystyle ,円\!{\begin{array}{lcl}{}_{p}\!\sum _{i=m}^{n}r^{i}&=&{r^{n+p} \over (r-1)^{p}}-\sum _{k=0}^{p-1}{r^{m+p-(k+1)}\prod _{j=1}^{k}(n-m+j) \over k!(r-1)^{p-k}}\\&=&{r^{n+p} \over (r-1)^{p}}-\sum _{k=0}^{p-1}{r^{m+p-(k+1)}{(n-m+k)! \over (n-m)!} \over (r-1)^{p-k}k!}\\&=&{r^{n+p} \over (r-1)^{p}}-\sum _{k=0}^{p-1}\left({r^{m+p-(k+1)} \over (r-1)^{p-k}}\right){{n-m+k} \choose k}\end{array}},円} {\displaystyle ,円\!{\begin{array}{lcl}{}_{p}\!\sum _{i=m}^{n}r^{i}&=&{r^{n+p} \over (r-1)^{p}}-\sum _{k=0}^{p-1}{r^{m+p-(k+1)}\prod _{j=1}^{k}(n-m+j) \over k!(r-1)^{p-k}}\\&=&{r^{n+p} \over (r-1)^{p}}-\sum _{k=0}^{p-1}{r^{m+p-(k+1)}{(n-m+k)! \over (n-m)!} \over (r-1)^{p-k}k!}\\&=&{r^{n+p} \over (r-1)^{p}}-\sum _{k=0}^{p-1}\left({r^{m+p-(k+1)} \over (r-1)^{p-k}}\right){{n-m+k} \choose k}\end{array}},円}

Combinations proof (used in below proof)

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F o r   n o n − n e g a t i v e   i n t e g e r s   q   a n d   t ≥ q , ∑ z = q t ( z q ) = ( t + 1 q + 1 ) . {\displaystyle \mathrm {For\ non-negative\ integers\ } q\ \mathrm {and\ } t\geq q,\quad \sum _{z=q}^{t}{z \choose q}={t+1 \choose q+1}.,円} {\displaystyle \mathrm {For\ non-negative\ integers\ } q\ \mathrm {and\ } t\geq q,\quad \sum _{z=q}^{t}{z \choose q}={t+1 \choose q+1}.,円}

t = q , ∑ z = q q ( z q ) = ( q q ) = 1 = ( q + 1 q + 1 ) {\displaystyle t=q,\quad \sum _{z=q}^{q}{z \choose q}={q \choose q}=1={q+1 \choose q+1},円} {\displaystyle t=q,\quad \sum _{z=q}^{q}{z \choose q}={q \choose q}=1={q+1 \choose q+1},円}

t = c , ∑ z = q c ( z q ) = ( c + 1 q + 1 ) {\displaystyle t=c,\quad \sum _{z=q}^{c}{z \choose q}={c+1 \choose q+1},円} {\displaystyle t=c,\quad \sum _{z=q}^{c}{z \choose q}={c+1 \choose q+1},円}

t = c + 1 , ∑ z = q c + 1 ( z q ) = ∑ z = q c ( z q ) + ( c + 1 q ) = ( c + 1 q + 1 ) + ( c + 1 q ) = ( ( c + 1 ) + 1 q + 1 ) {\displaystyle t=c+1,\quad \sum _{z=q}^{c+1}{z \choose q}=\sum _{z=q}^{c}{z \choose q}+{c+1 \choose q}={c+1 \choose q+1}+{c+1 \choose q}={(c+1)+1 \choose q+1},円} {\displaystyle t=c+1,\quad \sum _{z=q}^{c+1}{z \choose q}=\sum _{z=q}^{c}{z \choose q}+{c+1 \choose q}={c+1 \choose q+1}+{c+1 \choose q}={(c+1)+1 \choose q+1},円}

Proof by mathematical induction of the recursive geometric series (uses recursive summation notation)

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Definition

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F o r   a   r e a l   n u m b e r   r ≠ 1   a n d   n o n − n e g a t i v e   i n t e g e r s   m ,   p ,   a n d   i p ,   w h e r e   i p ≥ m , {\displaystyle \mathrm {For\ a\ real\ number\ } r\neq 1\ \mathrm {and\ non-negative\ integers\ } m,\ p,\ \mathrm {and\ } i_{p},\ \mathrm {where\ } i_{p}\geq m,} {\displaystyle \mathrm {For\ a\ real\ number\ } r\neq 1\ \mathrm {and\ non-negative\ integers\ } m,\ p,\ \mathrm {and\ } i_{p},\ \mathrm {where\ } i_{p}\geq m,}

∑ i = m i p p r i = r i p + p ( r − 1 ) p − ∑ k = 0 p − 1 r m + p − ( k + 1 ) ∏ j = 1 k ( i p − m + j ) k ! ( r − 1 ) p − k . {\displaystyle \sum _{i=m}^{i_{p}}\!\!{}^{p},円r^{i}={r^{i_{p}+p} \over (r-1)^{p}}-\sum _{k=0}^{p-1}{r^{m+p-(k+1)}\prod _{j=1}^{k}(i_{p}-m+j) \over k!(r-1)^{p-k}}.,円} {\displaystyle \sum _{i=m}^{i_{p}}\!\!{}^{p},円r^{i}={r^{i_{p}+p} \over (r-1)^{p}}-\sum _{k=0}^{p-1}{r^{m+p-(k+1)}\prod _{j=1}^{k}(i_{p}-m+j) \over k!(r-1)^{p-k}}.,円}

Base case (and some specific examples)

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p = 0 , ∑ i = m i 0 0 r i = r i 0 + 0 ( r − 1 ) 0 = r i 0 {\displaystyle p=0,\quad \sum _{i=m}^{i_{0}}\!\!{}^{0},円r^{i}={\cfrac {r^{i_{0}+0}}{(r-1)^{0}}}=r^{i_{0}},円} {\displaystyle p=0,\quad \sum _{i=m}^{i_{0}}\!\!{}^{0},円r^{i}={\cfrac {r^{i_{0}+0}}{(r-1)^{0}}}=r^{i_{0}},円}

p = 1 , ∑ i = m i 1 1 r i = r i 1 + 1 ( r − 1 ) 1 − r m 0 ! ( r − 1 ) 1 = r i 1 + 1 − r m r − 1 {\displaystyle p=1,\quad \sum _{i=m}^{i_{1}}\!\!{}^{1},円r^{i}={\cfrac {r^{i_{1}+1}}{(r-1)^{1}}}-{\cfrac {r^{m}}{0!(r-1)^{1}}}={\cfrac {r^{i_{1}+1}-r^{m}}{r-1}},円} {\displaystyle p=1,\quad \sum _{i=m}^{i_{1}}\!\!{}^{1},円r^{i}={\cfrac {r^{i_{1}+1}}{(r-1)^{1}}}-{\cfrac {r^{m}}{0!(r-1)^{1}}}={\cfrac {r^{i_{1}+1}-r^{m}}{r-1}},円}

p = 2 , ∑ i = m i 2 2 r i = r i 2 + 2 ( r − 1 ) 2 − r m + 1 0 ! ( r − 1 ) 2 − r m ( i 2 − m + 1 ) 1 ! ( r − 1 ) 1 = r i 2 + 2 − r m + 1 r − 1 − r m ( i 2 − m + 1 ) r − 1 {\displaystyle p=2,\quad \sum _{i=m}^{i_{2}}\!\!{}^{2},円r^{i}={\cfrac {r^{i_{2}+2}}{(r-1)^{2}}}-{\cfrac {r^{m+1}}{0!(r-1)^{2}}}-{\cfrac {r^{m}(i_{2}-m+1)}{1!(r-1)^{1}}}={\cfrac {{\cfrac {r^{i_{2}+2}-r^{m+1}}{r-1}}-r^{m}(i_{2}-m+1)}{r-1}},円} {\displaystyle p=2,\quad \sum _{i=m}^{i_{2}}\!\!{}^{2},円r^{i}={\cfrac {r^{i_{2}+2}}{(r-1)^{2}}}-{\cfrac {r^{m+1}}{0!(r-1)^{2}}}-{\cfrac {r^{m}(i_{2}-m+1)}{1!(r-1)^{1}}}={\cfrac {{\cfrac {r^{i_{2}+2}-r^{m+1}}{r-1}}-r^{m}(i_{2}-m+1)}{r-1}},円}

p = 3 , {\displaystyle p=3,,円} {\displaystyle p=3,,円}

∑ i = m i 3 3 r i = r i 3 + 3 ( r − 1 ) 3 − r m + 2 0 ! ( r − 1 ) 3 − r m + 1 ( i 3 − m + 1 ) 1 ! ( r − 1 ) 2 − r m ( i 3 − m + 1 ) ( i 3 − m + 2 ) 2 ! ( r − 1 ) 1 {\displaystyle \quad \sum _{i=m}^{i_{3}}\!\!{}^{3},円r^{i}={\cfrac {r^{i_{3}+3}}{(r-1)^{3}}}-{\cfrac {r^{m+2}}{0!(r-1)^{3}}}-{\cfrac {r^{m+1}(i_{3}-m+1)}{1!(r-1)^{2}}}-{\cfrac {r^{m}(i_{3}-m+1)(i_{3}-m+2)}{2!(r-1)^{1}}},円} {\displaystyle \quad \sum _{i=m}^{i_{3}}\!\!{}^{3},円r^{i}={\cfrac {r^{i_{3}+3}}{(r-1)^{3}}}-{\cfrac {r^{m+2}}{0!(r-1)^{3}}}-{\cfrac {r^{m+1}(i_{3}-m+1)}{1!(r-1)^{2}}}-{\cfrac {r^{m}(i_{3}-m+1)(i_{3}-m+2)}{2!(r-1)^{1}}},円}

= r i 3 + 3 − r m + 2 r − 1 − r m + 1 ( i 3 − m + 1 ) r − 1 − 1 2 r m ( i 3 − m + 1 ) ( i 3 − m + 2 ) r − 1 {\displaystyle ={\cfrac {{\cfrac {{\cfrac {r^{i_{3}+3}-r^{m+2}}{r-1}}-r^{m+1}(i_{3}-m+1)}{r-1}}-{\tfrac {1}{2}}r^{m}(i_{3}-m+1)(i_{3}-m+2)}{r-1}},円} {\displaystyle ={\cfrac {{\cfrac {{\cfrac {r^{i_{3}+3}-r^{m+2}}{r-1}}-r^{m+1}(i_{3}-m+1)}{r-1}}-{\tfrac {1}{2}}r^{m}(i_{3}-m+1)(i_{3}-m+2)}{r-1}},円}

p = 4 , {\displaystyle p=4,,円} {\displaystyle p=4,,円}

∑ i = m i 4 4 r i = r i 4 + 4 ( r − 1 ) 4 − r m + 3 0 ! ( r − 1 ) 4 − r m + 2 ( i 4 − m + 1 ) 1 ! ( r − 1 ) 3 − r m + 1 ( i 4 − m + 1 ) ( i 4 − m + 2 ) 2 ! ( r − 1 ) 2 {\displaystyle \quad \sum _{i=m}^{i_{4}}\!\!{}^{4},円r^{i}={\cfrac {r^{i_{4}+4}}{(r-1)^{4}}}-{\cfrac {r^{m+3}}{0!(r-1)^{4}}}-{\cfrac {r^{m+2}(i_{4}-m+1)}{1!(r-1)^{3}}}-{\cfrac {r^{m+1}(i_{4}-m+1)(i_{4}-m+2)}{2!(r-1)^{2}}},円} {\displaystyle \quad \sum _{i=m}^{i_{4}}\!\!{}^{4},円r^{i}={\cfrac {r^{i_{4}+4}}{(r-1)^{4}}}-{\cfrac {r^{m+3}}{0!(r-1)^{4}}}-{\cfrac {r^{m+2}(i_{4}-m+1)}{1!(r-1)^{3}}}-{\cfrac {r^{m+1}(i_{4}-m+1)(i_{4}-m+2)}{2!(r-1)^{2}}},円}

− r m ( i 4 − m + 1 ) ( i 4 − m + 2 ) ( i 4 − m + 3 ) 3 ! ( r − 1 ) 1 {\displaystyle -{\cfrac {r^{m}(i_{4}-m+1)(i_{4}-m+2)(i_{4}-m+3)}{3!(r-1)^{1}}},円} {\displaystyle -{\cfrac {r^{m}(i_{4}-m+1)(i_{4}-m+2)(i_{4}-m+3)}{3!(r-1)^{1}}},円}
= r i 4 + 4 − r m + 3 r − 1 − r m + 2 ( i 4 − m + 1 ) r − 1 − 1 2 r m + 1 ( i 4 − m + 1 ) ( i 4 − m + 2 ) r − 1 − 1 6 r m ( i 4 − m + 1 ) ( i 4 − m + 2 ) ( i 4 − m + 3 ) r − 1 {\displaystyle ={\cfrac {{\cfrac {{\cfrac {{\cfrac {r^{i_{4}+4}-r^{m+3}}{r-1}}-r^{m+2}(i_{4}-m+1)}{r-1}}-{\tfrac {1}{2}}r^{m+1}(i_{4}-m+1)(i_{4}-m+2)}{r-1}}-{\tfrac {1}{6}}r^{m}(i_{4}-m+1)(i_{4}-m+2)(i_{4}-m+3)}{r-1}},円} {\displaystyle ={\cfrac {{\cfrac {{\cfrac {{\cfrac {r^{i_{4}+4}-r^{m+3}}{r-1}}-r^{m+2}(i_{4}-m+1)}{r-1}}-{\tfrac {1}{2}}r^{m+1}(i_{4}-m+1)(i_{4}-m+2)}{r-1}}-{\tfrac {1}{6}}r^{m}(i_{4}-m+1)(i_{4}-m+2)(i_{4}-m+3)}{r-1}},円}

Inductive step

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p = a , ∑ i = m i a a r i = r i a + a ( r − 1 ) a − ∑ k = 0 a − 1 r m + a − ( k + 1 ) ∏ j = 1 k ( i a − m + j ) k ! ( r − 1 ) a − k {\displaystyle p=a,\quad \sum _{i=m}^{i_{a}}\!\!{}^{a},円r^{i}={r^{{i_{a}}+a} \over (r-1)^{a}}-\sum _{k=0}^{a-1}{r^{m+a-(k+1)}\prod _{j=1}^{k}(i_{a}-m+j) \over k!(r-1)^{a-k}},円} {\displaystyle p=a,\quad \sum _{i=m}^{i_{a}}\!\!{}^{a},円r^{i}={r^{{i_{a}}+a} \over (r-1)^{a}}-\sum _{k=0}^{a-1}{r^{m+a-(k+1)}\prod _{j=1}^{k}(i_{a}-m+j) \over k!(r-1)^{a-k}},円}

p = a + 1 , ∑ i = m i ( a + 1 ) ( a + 1 ) r i = ∑ i a = m i ( a + 1 ) [ ∑ i = m i a a r i ] {\displaystyle p=a+1,\quad \sum _{i=m}^{i_{(a+1)}}\!\!{}^{(a+1)},円r^{i}=\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[\sum _{i=m}^{i_{a}}\!\!{}^{a},円r^{i}\right],円} {\displaystyle p=a+1,\quad \sum _{i=m}^{i_{(a+1)}}\!\!{}^{(a+1)},円r^{i}=\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[\sum _{i=m}^{i_{a}}\!\!{}^{a},円r^{i}\right],円}

= ∑ i a = m i ( a + 1 ) [ r i a + a ( r − 1 ) a − ∑ k = 0 a − 1 r m + a − ( k + 1 ) ∏ j = 1 k ( i a − m + j ) k ! ( r − 1 ) a − k ] {\displaystyle =\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[{r^{{i_{a}}+a} \over (r-1)^{a}}-\sum _{k=0}^{a-1}{r^{m+a-(k+1)}\prod _{j=1}^{k}(i_{a}-m+j) \over k!(r-1)^{a-k}}\right],円} {\displaystyle =\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[{r^{{i_{a}}+a} \over (r-1)^{a}}-\sum _{k=0}^{a-1}{r^{m+a-(k+1)}\prod _{j=1}^{k}(i_{a}-m+j) \over k!(r-1)^{a-k}}\right],円}

= ∑ i a = m i ( a + 1 ) [ r i a + a ( r − 1 ) a ] − ∑ i a = m i ( a + 1 ) [ ∑ k = 0 a − 1 r m + a − ( k + 1 ) ∏ j = 1 k ( i a − m + j ) k ! ( r − 1 ) a − k ] {\displaystyle =\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[{r^{{i_{a}}+a} \over (r-1)^{a}}\right]-\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[\sum _{k=0}^{a-1}{r^{m+a-(k+1)}\prod _{j=1}^{k}(i_{a}-m+j) \over k!(r-1)^{a-k}}\right],円} {\displaystyle =\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[{r^{{i_{a}}+a} \over (r-1)^{a}}\right]-\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[\sum _{k=0}^{a-1}{r^{m+a-(k+1)}\prod _{j=1}^{k}(i_{a}-m+j) \over k!(r-1)^{a-k}}\right],円}

= ∑ i a = m i ( a + 1 ) ( r i a + a ) ( r − 1 ) a − ∑ i a = m i ( a + 1 ) [ ∑ k = 0 a − 1 r m + a − ( k + 1 ) [ ( i a − m + k ) ! ( i a − m ) ! ] ( r − 1 ) a − k k ! ] {\displaystyle ={\sum _{{i_{a}}=m}^{i_{(a+1)}}\left(r^{{i_{a}}+a}\right) \over (r-1)^{a}}-\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[\sum _{k=0}^{a-1}{r^{m+a-(k+1)}\left[{(i_{a}-m+k)! \over (i_{a}-m)!}\right] \over (r-1)^{a-k}\qquad k!}\right],円} {\displaystyle ={\sum _{{i_{a}}=m}^{i_{(a+1)}}\left(r^{{i_{a}}+a}\right) \over (r-1)^{a}}-\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[\sum _{k=0}^{a-1}{r^{m+a-(k+1)}\left[{(i_{a}-m+k)! \over (i_{a}-m)!}\right] \over (r-1)^{a-k}\qquad k!}\right],円}

= r a ∑ i a = m i ( a + 1 ) ( r i a ) ( r − 1 ) a − ∑ i a = m i ( a + 1 ) [ ∑ k = 0 a − 1 ( r m + a − ( k + 1 ) ( r − 1 ) a − k ) ( i a − m + k k ) ] {\displaystyle ={r^{a}\sum _{{i_{a}}=m}^{i_{(a+1)}}\left(r^{i_{a}}\right) \over (r-1)^{a}}-\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[\sum _{k=0}^{a-1}\left({r^{m+a-(k+1)} \over (r-1)^{a-k}}\right){i_{a}-m+k \choose k}\right],円} {\displaystyle ={r^{a}\sum _{{i_{a}}=m}^{i_{(a+1)}}\left(r^{i_{a}}\right) \over (r-1)^{a}}-\sum _{{i_{a}}=m}^{i_{(a+1)}}\left[\sum _{k=0}^{a-1}\left({r^{m+a-(k+1)} \over (r-1)^{a-k}}\right){i_{a}-m+k \choose k}\right],円}

= r a ( r i ( a + 1 ) + 1 − r m r − 1 ) ( r − 1 ) a − ∑ k = 0 a − 1 [ ( r m + a − ( k + 1 ) ( r − 1 ) a − k ) ∑ i a = m i ( a + 1 ) ( i a − m + k k ) ] {\displaystyle ={r^{a}\left({r^{i_{(a+1)}+1}-r^{m} \over r-1}\right) \over (r-1)^{a}}-\sum _{k=0}^{a-1}\left[\left({r^{m+a-(k+1)} \over (r-1)^{a-k}}\right)\sum _{{i_{a}}=m}^{i_{(a+1)}}{i_{a}-m+k \choose k}\right],円} {\displaystyle ={r^{a}\left({r^{i_{(a+1)}+1}-r^{m} \over r-1}\right) \over (r-1)^{a}}-\sum _{k=0}^{a-1}\left[\left({r^{m+a-(k+1)} \over (r-1)^{a-k}}\right)\sum _{{i_{a}}=m}^{i_{(a+1)}}{i_{a}-m+k \choose k}\right],円}

Shifting of starting and ending indices (see above for proof):

= r i ( a + 1 ) + a + 1 − r m + a ( r − 1 ) a + 1 − ∑ k = 0 a − 1 [ ( r m + a − ( k + 1 ) ( r − 1 ) a − k ) ∑ i a = k i ( a + 1 ) − m + k ( i a k ) ] {\displaystyle ={r^{i_{(a+1)}+a+1}-r^{m+a} \over (r-1)^{a+1}}-\sum _{k=0}^{a-1}\left[\left({r^{m+a-(k+1)} \over (r-1)^{a-k}}\right)\sum _{i_{a}=k}^{i_{(a+1)}-m+k}{i_{a} \choose k}\right],円} {\displaystyle ={r^{i_{(a+1)}+a+1}-r^{m+a} \over (r-1)^{a+1}}-\sum _{k=0}^{a-1}\left[\left({r^{m+a-(k+1)} \over (r-1)^{a-k}}\right)\sum _{i_{a}=k}^{i_{(a+1)}-m+k}{i_{a} \choose k}\right],円}

See combinations proof above:

= r i ( a + 1 ) + a + 1 ( r − 1 ) a + 1 − r m + a ( r − 1 ) a + 1 − ∑ k = 0 a − 1 ( r m + a − ( k + 1 ) ( r − 1 ) a − k ) ( i ( a + 1 ) − m + k + 1 k + 1 ) {\displaystyle ={r^{i_{(a+1)}+a+1} \over (r-1)^{a+1}}-{r^{m+a} \over (r-1)^{a+1}}-\sum _{k=0}^{a-1}\left({r^{m+a-(k+1)} \over (r-1)^{a-k}}\right){i_{(a+1)}-m+k+1 \choose k+1},円} {\displaystyle ={r^{i_{(a+1)}+a+1} \over (r-1)^{a+1}}-{r^{m+a} \over (r-1)^{a+1}}-\sum _{k=0}^{a-1}\left({r^{m+a-(k+1)} \over (r-1)^{a-k}}\right){i_{(a+1)}-m+k+1 \choose k+1},円}

Shifting of starting and ending indices (see above for proof):

= r i ( a + 1 ) + a + 1 ( r − 1 ) a + 1 − r m + a ( r − 1 ) a + 1 − ∑ k = 1 ( a + 1 ) − 1 ( r m + a − ( ( k − 1 ) + 1 ) ( r − 1 ) a − ( k − 1 ) ) ( i ( a + 1 ) − m + ( k − 1 ) + 1 ( k − 1 ) + 1 ) {\displaystyle ={r^{i_{(a+1)}+a+1} \over (r-1)^{a+1}}-{r^{m+a} \over (r-1)^{a+1}}-\sum _{k=1}^{(a+1)-1}\left({r^{m+a-((k-1)+1)} \over (r-1)^{a-(k-1)}}\right){i_{(a+1)}-m+(k-1)+1 \choose (k-1)+1},円} {\displaystyle ={r^{i_{(a+1)}+a+1} \over (r-1)^{a+1}}-{r^{m+a} \over (r-1)^{a+1}}-\sum _{k=1}^{(a+1)-1}\left({r^{m+a-((k-1)+1)} \over (r-1)^{a-(k-1)}}\right){i_{(a+1)}-m+(k-1)+1 \choose (k-1)+1},円}

= r i ( a + 1 ) + a + 1 ( r − 1 ) a + 1 − r m + a ( r − 1 ) a + 1 − ∑ k = 1 ( a + 1 ) − 1 ( r m + a − k + 1 − 1 ( r − 1 ) a − k + 1 ) ( i ( a + 1 ) − m + k k ) {\displaystyle ={r^{i_{(a+1)}+a+1} \over (r-1)^{a+1}}-{r^{m+a} \over (r-1)^{a+1}}-\sum _{k=1}^{(a+1)-1}\left({r^{m+a-k+1-1} \over (r-1)^{a-k+1}}\right){i_{(a+1)}-m+k \choose k},円} {\displaystyle ={r^{i_{(a+1)}+a+1} \over (r-1)^{a+1}}-{r^{m+a} \over (r-1)^{a+1}}-\sum _{k=1}^{(a+1)-1}\left({r^{m+a-k+1-1} \over (r-1)^{a-k+1}}\right){i_{(a+1)}-m+k \choose k},円}

Adding case k=0 to the summation, means that the same must be subtracted from the summation:

= r i ( a + 1 ) + ( a + 1 ) ( r − 1 ) ( a + 1 ) − r m + a ( r − 1 ) a + 1 − [ ∑ k = 0 ( a + 1 ) − 1 ( r m + ( a + 1 ) − k − 1 ( r − 1 ) ( a + 1 ) − k ) ( i ( a + 1 ) − m + k k ) − ( r m + a ( r − 1 ) a + 1 ) ( i ( a + 1 ) − m 0 ) ] {\displaystyle ={r^{i_{(a+1)}+(a+1)} \over (r-1)^{(a+1)}}-{r^{m+a} \over (r-1)^{a+1}}-\left[\sum _{k=0}^{(a+1)-1}\left({r^{m+(a+1)-k-1} \over (r-1)^{(a+1)-k}}\right){i_{(a+1)}-m+k \choose k}-\left({r^{m+a} \over (r-1)^{a+1}}\right){i_{(a+1)}-m \choose 0}\right],円} {\displaystyle ={r^{i_{(a+1)}+(a+1)} \over (r-1)^{(a+1)}}-{r^{m+a} \over (r-1)^{a+1}}-\left[\sum _{k=0}^{(a+1)-1}\left({r^{m+(a+1)-k-1} \over (r-1)^{(a+1)-k}}\right){i_{(a+1)}-m+k \choose k}-\left({r^{m+a} \over (r-1)^{a+1}}\right){i_{(a+1)}-m \choose 0}\right],円}

Terms cancel out.

= r i ( a + 1 ) + ( a + 1 ) ( r − 1 ) ( a + 1 ) − ∑ k = 0 ( a + 1 ) − 1 ( r m + ( a + 1 ) − ( k + 1 ) ( r − 1 ) ( a + 1 ) − k ) ( i ( a + 1 ) − m + k k ) {\displaystyle ={r^{i_{(a+1)}+(a+1)} \over (r-1)^{(a+1)}}-\sum _{k=0}^{(a+1)-1}\left({r^{m+(a+1)-(k+1)} \over (r-1)^{(a+1)-k}}\right){i_{(a+1)}-m+k \choose k},円} {\displaystyle ={r^{i_{(a+1)}+(a+1)} \over (r-1)^{(a+1)}}-\sum _{k=0}^{(a+1)-1}\left({r^{m+(a+1)-(k+1)} \over (r-1)^{(a+1)-k}}\right){i_{(a+1)}-m+k \choose k},円}

= r i ( a + 1 ) + ( a + 1 ) ( r − 1 ) ( a + 1 ) − ∑ k = 0 ( a + 1 ) − 1 r m + ( a + 1 ) − ( k + 1 ) [ ( i ( a + 1 ) − m + k ) ! ( i ( a + 1 ) − m ) ! ] ( r − 1 ) ( a + 1 ) − k k ! {\displaystyle ={r^{i_{(a+1)}+(a+1)} \over (r-1)^{(a+1)}}-\sum _{k=0}^{(a+1)-1}{r^{m+(a+1)-(k+1)}\left[{(i_{(a+1)}-m+k)! \over (i_{(a+1)}-m)!}\right] \over (r-1)^{(a+1)-k}\qquad k!},円} {\displaystyle ={r^{i_{(a+1)}+(a+1)} \over (r-1)^{(a+1)}}-\sum _{k=0}^{(a+1)-1}{r^{m+(a+1)-(k+1)}\left[{(i_{(a+1)}-m+k)! \over (i_{(a+1)}-m)!}\right] \over (r-1)^{(a+1)-k}\qquad k!},円}

∑ i = m i ( a + 1 ) ( a + 1 ) r i = r i ( a + 1 ) + ( a + 1 ) ( r − 1 ) ( a + 1 ) − ∑ k = 0 ( a + 1 ) − 1 r m + ( a + 1 ) − ( k + 1 ) ∏ j = 1 k ( i ( a + 1 ) − m + j ) k ! ( r − 1 ) ( a + 1 ) − k {\displaystyle \sum _{i=m}^{i_{(a+1)}}\!\!{}^{(a+1)},円r^{i}={r^{i_{(a+1)}+(a+1)} \over (r-1)^{(a+1)}}-\sum _{k=0}^{(a+1)-1}{r^{m+(a+1)-(k+1)}\prod _{j=1}^{k}(i_{(a+1)}-m+j) \over k!(r-1)^{(a+1)-k}},円} {\displaystyle \sum _{i=m}^{i_{(a+1)}}\!\!{}^{(a+1)},円r^{i}={r^{i_{(a+1)}+(a+1)} \over (r-1)^{(a+1)}}-\sum _{k=0}^{(a+1)-1}{r^{m+(a+1)-(k+1)}\prod _{j=1}^{k}(i_{(a+1)}-m+j) \over k!(r-1)^{(a+1)-k}},円}

Q.E.D.

A general formula for recursive summation series

[edit ]

First proof, used in second proof

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One method

[edit ]

n ≥ m , ∑ i = m n [ ∑ j = m i ( i − j + k k ) f ( j ) ] = ∑ j = m m ( ( m ) − j + k k ) f ( j ) + ∑ j = m m + 1 ( ( m + 1 ) − j + k k ) f ( j ) + ∑ j = m m + 2 ( ( m + 2 ) − j + k k ) f ( j ) + ⋯ + ∑ j = m n − 2 ( ( n − 2 ) − j + k k ) f ( j ) + ∑ j = m n − 1 ( ( n − 1 ) − j + k k ) f ( j ) + ∑ j = m n ( ( n ) − j + k k ) f ( j ) = [ ( ( m ) − ( m ) + k k ) f ( m ) ] + [ ( ( m + 1 ) − ( m ) + k k ) f ( m ) + ( ( m + 1 ) − ( m + 1 ) + k k ) f ( m + 1 ) ] + [ ( ( m + 2 ) − ( m ) + k k ) f ( m ) + ( ( m + 2 ) − ( m + 1 ) + k k ) f ( m + 1 ) + ( ( m + 2 ) − ( m + 2 ) + k k ) f ( m + 2 ) ] + ⋯ + [ ( ( n − 2 ) − ( m ) + k k ) f ( m ) + ( ( n − 2 ) − ( m + 1 ) + k k ) f ( m + 1 ) + ( ( n − 2 ) − ( m + 2 ) + k k ) f ( m + 2 ) + ⋯ + ( ( n − 2 ) − ( n − 4 ) + k k ) f ( n − 4 ) + ( ( n − 2 ) − ( n − 3 ) + k k ) f ( n − 3 ) + ( ( n − 2 ) − ( n − 2 ) + k k ) f ( n − 2 ) ] + [ ( ( n − 1 ) − ( m ) + k k ) f ( m ) + ( ( n − 1 ) − ( m + 1 ) + k k ) f ( m + 1 ) + ( ( n − 1 ) − ( m + 2 ) + k k ) f ( m + 2 ) + ⋯ + ( ( n − 1 ) − ( n − 3 ) + k k ) f ( n − 3 ) + ( ( n − 1 ) − ( n − 2 ) + k k ) f ( n − 2 ) + ( ( n − 1 ) − ( n − 1 ) + k k ) f ( n − 1 ) ] + [ ( ( n ) − ( m ) + k k ) f ( m ) + ( ( n ) − ( m + 1 ) + k k ) f ( m + 1 ) + ( ( n ) − ( m + 2 ) + k k ) f ( m + 2 ) + ⋯ + ( ( n ) − ( n − 2 ) + k k ) f ( n − 2 ) + ( ( n ) − ( n − 1 ) + k k ) f ( n − 1 ) + ( ( n ) − ( n ) + k k ) f ( n ) ] = [ ( k k ) f ( m ) ] + [ ( 1 + k k ) f ( m ) + ( k k ) f ( m + 1 ) ] + [ ( 2 + k k ) f ( m ) + ( 1 + k k ) f ( m + 1 ) + ( k k ) f ( m ) ] + ⋯ + [ ( n − 2 − m + k k ) f ( m ) + ( n − 3 − m + k k ) f ( m + 1 ) + ( n − 4 − m + k k ) f ( m + 2 ) + ⋯ + ( 2 + k k ) f ( n − 4 ) + ( 1 + k k ) f ( n − 3 ) + ( k k ) f ( n − 2 ) ] + [ ( n − 1 − m + k k ) f ( m ) + ( n − 2 − m + k k ) f ( m + 1 ) + ( n − 3 − m + k k ) f ( m + 2 ) + ⋯ + ( 2 + k k ) f ( n − 3 ) + ( 1 + k k ) f ( n − 2 ) + ( k k ) f ( n − 1 ) ] + [ ( n − m + k k ) f ( m ) + ( n − 1 − m + k k ) f ( m + 1 ) + ( n − 2 − m + k k ) f ( m + 2 ) + ⋯ + ( 2 + k k ) f ( n − 2 ) + ( 1 + k k ) f ( n − 1 ) + ( k k ) f ( n ) ] = f ( m ) [ ( k k ) + ( 1 + k k ) + ( 2 + k k ) + ⋯ + ( n − 2 − m + k k ) + ( n − 1 − m + k k ) + ( n − m + k k ) ] + f ( m + 1 ) [ ( k k ) + ( 1 + k k ) + ( 2 + k k ) + ⋯ + ( n − 3 − m + k k ) + ( n − 2 − m + k k ) + ( n − 1 − m + k k ) ] + f ( m + 2 ) [ ( k k ) + ( 1 + k k ) + ( 2 + k k ) + ⋯ + ( n − 4 − m + k k ) + ( n − 3 − m + k k ) + ( n − 2 − m + k k ) ] + ⋯ + f ( n − 2 ) [ ( k k ) + ( 1 + k k ) + ( 2 + k k ) ] + f ( n − 1 ) [ ( k k ) + ( 1 + k k ) ] + f ( n ) [ ( k k ) ] = ∑ j = m n ( n − j + k + 1 k + 1 ) f ( j ) {\displaystyle {\begin{aligned}n\geq m,&\sum _{i=m}^{n}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{m}{(m)-j+k \choose k}f(j)+\sum _{j=m}^{m+1}{(m+1)-j+k \choose k}f(j)\\&&&+\sum _{j=m}^{m+2}{(m+2)-j+k \choose k}f(j)+\cdots +\sum _{j=m}^{n-2}{(n-2)-j+k \choose k}f(j)\\&&&+\sum _{j=m}^{n-1}{(n-1)-j+k \choose k}f(j)+\sum _{j=m}^{n}{(n)-j+k \choose k}f(j)\\&&=&\left[{(m)-(m)+k \choose k}f(m)\right]+\left[{(m+1)-(m)+k \choose k}f(m)\right.\\&&&\left.+{(m+1)-(m+1)+k \choose k}f(m+1)\right]+\left[{(m+2)-(m)+k \choose k}f(m)\right.\\&&&\left.+{(m+2)-(m+1)+k \choose k}f(m+1)+{(m+2)-(m+2)+k \choose k}f(m+2)\right]\\&&&+\cdots +\left[{(n-2)-(m)+k \choose k}f(m)+{(n-2)-(m+1)+k \choose k}f(m+1)\right.\\&&&\left.+{(n-2)-(m+2)+k \choose k}f(m+2)+\cdots +{(n-2)-(n-4)+k \choose k}f(n-4)\right.\\&&&\left.+{(n-2)-(n-3)+k \choose k}f(n-3)+{(n-2)-(n-2)+k \choose k}f(n-2)\right]\\&&&+\left[{(n-1)-(m)+k \choose k}f(m)+{(n-1)-(m+1)+k \choose k}f(m+1)\right.\\&&&\left.+{(n-1)-(m+2)+k \choose k}f(m+2)+\cdots +{(n-1)-(n-3)+k \choose k}f(n-3)\right.\\&&&\left.+{(n-1)-(n-2)+k \choose k}f(n-2)+{(n-1)-(n-1)+k \choose k}f(n-1)\right]\\&&&+\left[{(n)-(m)+k \choose k}f(m)+{(n)-(m+1)+k \choose k}f(m+1)\right.\\&&&\left.+{(n)-(m+2)+k \choose k}f(m+2)+\cdots +{(n)-(n-2)+k \choose k}f(n-2)\right.\\&&&\left.+{(n)-(n-1)+k \choose k}f(n-1)+{(n)-(n)+k \choose k}f(n)\right]\\&&=&\left[{k \choose k}f(m)\right]+\left[{1+k \choose k}f(m)+{k \choose k}f(m+1)\right]\\&&&+\left[{2+k \choose k}f(m)+{1+k \choose k}f(m+1)+{k \choose k}f(m)\right]\\&&&+\cdots +\left[{n-2-m+k \choose k}f(m)+{n-3-m+k \choose k}f(m+1)\right.\\&&&\left.+{n-4-m+k \choose k}f(m+2)+\cdots +{2+k \choose k}f(n-4)\right.\\&&&\left.+{1+k \choose k}f(n-3)+{k \choose k}f(n-2)\right]\\&&&+\left[{n-1-m+k \choose k}f(m)+{n-2-m+k \choose k}f(m+1)\right.\\&&&\left.+{n-3-m+k \choose k}f(m+2)+\cdots +{2+k \choose k}f(n-3)\right.\\&&&\left.+{1+k \choose k}f(n-2)+{k \choose k}f(n-1)\right]\\&&&+\left[{n-m+k \choose k}f(m)+{n-1-m+k \choose k}f(m+1)\right.\\&&&\left.+{n-2-m+k \choose k}f(m+2)+\cdots +{2+k \choose k}f(n-2)\right.\\&&&\left.+{1+k \choose k}f(n-1)+{k \choose k}f(n)\right]\\&&=&f(m)\left[{k \choose k}+{1+k \choose k}+{2+k \choose k}\right.\\&&&\left.+\cdots +{n-2-m+k \choose k}+{n-1-m+k \choose k}+{n-m+k \choose k}\right]\\&&&+f(m+1)\left[{k \choose k}+{1+k \choose k}+{2+k \choose k}\right.\\&&&\left.+\cdots +{n-3-m+k \choose k}+{n-2-m+k \choose k}+{n-1-m+k \choose k}\right]\\&&&+f(m+2)\left[{k \choose k}+{1+k \choose k}+{2+k \choose k}\right.\\&&&\left.+\cdots +{n-4-m+k \choose k}+{n-3-m+k \choose k}+{n-2-m+k \choose k}\right]\\&&&+\cdots +f(n-2)\left[{k \choose k}+{1+k \choose k}+{2+k \choose k}\right]\\&&&+f(n-1)\left[{k \choose k}+{1+k \choose k}\right]+f(n)\left[{k \choose k}\right]\\&&=&\sum _{j=m}^{n}{n-j+k+1 \choose k+1}f(j)\end{aligned}},円} {\displaystyle {\begin{aligned}n\geq m,&\sum _{i=m}^{n}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{m}{(m)-j+k \choose k}f(j)+\sum _{j=m}^{m+1}{(m+1)-j+k \choose k}f(j)\\&&&+\sum _{j=m}^{m+2}{(m+2)-j+k \choose k}f(j)+\cdots +\sum _{j=m}^{n-2}{(n-2)-j+k \choose k}f(j)\\&&&+\sum _{j=m}^{n-1}{(n-1)-j+k \choose k}f(j)+\sum _{j=m}^{n}{(n)-j+k \choose k}f(j)\\&&=&\left[{(m)-(m)+k \choose k}f(m)\right]+\left[{(m+1)-(m)+k \choose k}f(m)\right.\\&&&\left.+{(m+1)-(m+1)+k \choose k}f(m+1)\right]+\left[{(m+2)-(m)+k \choose k}f(m)\right.\\&&&\left.+{(m+2)-(m+1)+k \choose k}f(m+1)+{(m+2)-(m+2)+k \choose k}f(m+2)\right]\\&&&+\cdots +\left[{(n-2)-(m)+k \choose k}f(m)+{(n-2)-(m+1)+k \choose k}f(m+1)\right.\\&&&\left.+{(n-2)-(m+2)+k \choose k}f(m+2)+\cdots +{(n-2)-(n-4)+k \choose k}f(n-4)\right.\\&&&\left.+{(n-2)-(n-3)+k \choose k}f(n-3)+{(n-2)-(n-2)+k \choose k}f(n-2)\right]\\&&&+\left[{(n-1)-(m)+k \choose k}f(m)+{(n-1)-(m+1)+k \choose k}f(m+1)\right.\\&&&\left.+{(n-1)-(m+2)+k \choose k}f(m+2)+\cdots +{(n-1)-(n-3)+k \choose k}f(n-3)\right.\\&&&\left.+{(n-1)-(n-2)+k \choose k}f(n-2)+{(n-1)-(n-1)+k \choose k}f(n-1)\right]\\&&&+\left[{(n)-(m)+k \choose k}f(m)+{(n)-(m+1)+k \choose k}f(m+1)\right.\\&&&\left.+{(n)-(m+2)+k \choose k}f(m+2)+\cdots +{(n)-(n-2)+k \choose k}f(n-2)\right.\\&&&\left.+{(n)-(n-1)+k \choose k}f(n-1)+{(n)-(n)+k \choose k}f(n)\right]\\&&=&\left[{k \choose k}f(m)\right]+\left[{1+k \choose k}f(m)+{k \choose k}f(m+1)\right]\\&&&+\left[{2+k \choose k}f(m)+{1+k \choose k}f(m+1)+{k \choose k}f(m)\right]\\&&&+\cdots +\left[{n-2-m+k \choose k}f(m)+{n-3-m+k \choose k}f(m+1)\right.\\&&&\left.+{n-4-m+k \choose k}f(m+2)+\cdots +{2+k \choose k}f(n-4)\right.\\&&&\left.+{1+k \choose k}f(n-3)+{k \choose k}f(n-2)\right]\\&&&+\left[{n-1-m+k \choose k}f(m)+{n-2-m+k \choose k}f(m+1)\right.\\&&&\left.+{n-3-m+k \choose k}f(m+2)+\cdots +{2+k \choose k}f(n-3)\right.\\&&&\left.+{1+k \choose k}f(n-2)+{k \choose k}f(n-1)\right]\\&&&+\left[{n-m+k \choose k}f(m)+{n-1-m+k \choose k}f(m+1)\right.\\&&&\left.+{n-2-m+k \choose k}f(m+2)+\cdots +{2+k \choose k}f(n-2)\right.\\&&&\left.+{1+k \choose k}f(n-1)+{k \choose k}f(n)\right]\\&&=&f(m)\left[{k \choose k}+{1+k \choose k}+{2+k \choose k}\right.\\&&&\left.+\cdots +{n-2-m+k \choose k}+{n-1-m+k \choose k}+{n-m+k \choose k}\right]\\&&&+f(m+1)\left[{k \choose k}+{1+k \choose k}+{2+k \choose k}\right.\\&&&\left.+\cdots +{n-3-m+k \choose k}+{n-2-m+k \choose k}+{n-1-m+k \choose k}\right]\\&&&+f(m+2)\left[{k \choose k}+{1+k \choose k}+{2+k \choose k}\right.\\&&&\left.+\cdots +{n-4-m+k \choose k}+{n-3-m+k \choose k}+{n-2-m+k \choose k}\right]\\&&&+\cdots +f(n-2)\left[{k \choose k}+{1+k \choose k}+{2+k \choose k}\right]\\&&&+f(n-1)\left[{k \choose k}+{1+k \choose k}\right]+f(n)\left[{k \choose k}\right]\\&&=&\sum _{j=m}^{n}{n-j+k+1 \choose k+1}f(j)\end{aligned}},円}

Inductive method

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n ≥ m , ∑ i = m n [ ∑ j = m i ( i − j + k k ) f ( j ) ] = ∑ j = m n ( n − j + k + 1 k + 1 ) f ( j ) n = m , ∑ i = m m [ ∑ j = m i ( i − j + k k ) f ( j ) ] = ∑ j = m m ( ( m ) − j + k k ) f ( j ) = ( m − ( m ) + k k ) f ( m ) = ( k k ) f ( m ) = f ( m ) = ( k + 1 k + 1 ) f ( m ) = ( ( m ) − ( m ) + k + 1 k + 1 ) f ( m ) = ∑ j = m m ( ( m ) − j + k + 1 k + 1 ) f ( j ) n = x , ∑ i = m x [ ∑ j = m i ( i − j + k k ) f ( j ) ] = ∑ j = m x ( x − j + k + 1 k + 1 ) f ( j ) n = x + 1 , ∑ i = m x + 1 [ ∑ j = m i ( i − j + k k ) f ( j ) ] = ∑ j = m x + 1 ( ( x + 1 ) − j + k k ) f ( j ) + ∑ i = m x [ ∑ j = m i ( i − j + k k ) f ( j ) ] = ∑ j = m x + 1 ( ( x + 1 ) − j + k k ) f ( j ) + ∑ j = m x ( x − j + k + 1 k + 1 ) f ( j ) = ( ( x + 1 ) − ( x + 1 ) + k k ) f ( x + 1 ) + ∑ j = m x ( ( x + 1 ) − j + k k ) f ( j ) + ∑ j = m x ( x − j + k + 1 k + 1 ) f ( j ) = f ( x + 1 ) + ∑ j = m x ( x − j + k + 1 k ) f ( j ) + ∑ j = m x ( x − j + k + 1 k + 1 ) f ( j ) = f ( x + 1 ) + ∑ j = m x [ ( x − j + k + 1 k ) + ( x − j + k + 1 k + 1 ) ] f ( j ) = f ( x + 1 ) + ∑ j = m x ( x − j + k + 2 k + 1 ) f ( j ) = ( ( x + 1 ) − ( x + 1 ) + k + 1 k + 1 ) f ( x + 1 ) + ∑ j = m x ( ( x + 1 ) − j + k + 1 k + 1 ) f ( j ) = ∑ j = m x + 1 ( ( x + 1 ) − j + k + 1 k + 1 ) f ( j ) {\displaystyle {\begin{aligned}n\geq m,\quad &\sum _{i=m}^{n}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{n}{n-j+k+1 \choose k+1}f(j)\\n=m,\quad &\sum _{i=m}^{m}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{m}{(m)-j+k \choose k}f(j)\\&&=&{m-(m)+k \choose k}f(m)={k \choose k}f(m)=f(m)={k+1 \choose k+1}f(m)\\&&=&{(m)-(m)+k+1 \choose k+1}f(m)=\sum _{j=m}^{m}{(m)-j+k+1 \choose k+1}f(j)\\n=x,\quad &\sum _{i=m}^{x}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{x}{x-j+k+1 \choose k+1}f(j)\\n=x+1,\quad &\sum _{i=m}^{x+1}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{x+1}{(x+1)-j+k \choose k}f(j)+\sum _{i=m}^{x}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]\\&&=&\sum _{j=m}^{x+1}{(x+1)-j+k \choose k}f(j)+\sum _{j=m}^{x}{x-j+k+1 \choose k+1}f(j)\\&&=&{(x+1)-(x+1)+k \choose k}f(x+1)\\&&&+\sum _{j=m}^{x}{(x+1)-j+k \choose k}f(j)+\sum _{j=m}^{x}{x-j+k+1 \choose k+1}f(j)\\&&=&f(x+1)+\sum _{j=m}^{x}{x-j+k+1 \choose k}f(j)+\sum _{j=m}^{x}{x-j+k+1 \choose k+1}f(j)\\&&=&f(x+1)+\sum _{j=m}^{x}\left[{x-j+k+1 \choose k}+{x-j+k+1 \choose k+1}\right]f(j)\\&&=&f(x+1)+\sum _{j=m}^{x}{x-j+k+2 \choose k+1}f(j)\\&&=&{(x+1)-(x+1)+k+1 \choose k+1}f(x+1)+\sum _{j=m}^{x}{(x+1)-j+k+1 \choose k+1}f(j)\\&&=&\sum _{j=m}^{x+1}{(x+1)-j+k+1 \choose k+1}f(j)\end{aligned}},円} {\displaystyle {\begin{aligned}n\geq m,\quad &\sum _{i=m}^{n}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{n}{n-j+k+1 \choose k+1}f(j)\\n=m,\quad &\sum _{i=m}^{m}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{m}{(m)-j+k \choose k}f(j)\\&&=&{m-(m)+k \choose k}f(m)={k \choose k}f(m)=f(m)={k+1 \choose k+1}f(m)\\&&=&{(m)-(m)+k+1 \choose k+1}f(m)=\sum _{j=m}^{m}{(m)-j+k+1 \choose k+1}f(j)\\n=x,\quad &\sum _{i=m}^{x}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{x}{x-j+k+1 \choose k+1}f(j)\\n=x+1,\quad &\sum _{i=m}^{x+1}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]&=&\sum _{j=m}^{x+1}{(x+1)-j+k \choose k}f(j)+\sum _{i=m}^{x}\left[\sum _{j=m}^{i}{i-j+k \choose k}f(j)\right]\\&&=&\sum _{j=m}^{x+1}{(x+1)-j+k \choose k}f(j)+\sum _{j=m}^{x}{x-j+k+1 \choose k+1}f(j)\\&&=&{(x+1)-(x+1)+k \choose k}f(x+1)\\&&&+\sum _{j=m}^{x}{(x+1)-j+k \choose k}f(j)+\sum _{j=m}^{x}{x-j+k+1 \choose k+1}f(j)\\&&=&f(x+1)+\sum _{j=m}^{x}{x-j+k+1 \choose k}f(j)+\sum _{j=m}^{x}{x-j+k+1 \choose k+1}f(j)\\&&=&f(x+1)+\sum _{j=m}^{x}\left[{x-j+k+1 \choose k}+{x-j+k+1 \choose k+1}\right]f(j)\\&&=&f(x+1)+\sum _{j=m}^{x}{x-j+k+2 \choose k+1}f(j)\\&&=&{(x+1)-(x+1)+k+1 \choose k+1}f(x+1)+\sum _{j=m}^{x}{(x+1)-j+k+1 \choose k+1}f(j)\\&&=&\sum _{j=m}^{x+1}{(x+1)-j+k+1 \choose k+1}f(j)\end{aligned}},円}

Second proof, this one for the general formula for recursive summation series

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p ≥ 1 , ∑ i = m i p p   f ( i ) = ∑ i = m i p ( i p − i + p − 1 p − 1 ) f ( i ) p = 1 , ∑ i = m i 1 1   f ( i ) = ∑ i = m i 1 ( i 1 − i 0 ) f ( i ) = ∑ i = m i 1 f ( i ) p = s , ∑ i = m i s s   f ( i ) = ∑ i = m i s ( i s − i + s − 1 s − 1 ) f ( i ) p = s + 1 , ∑ i = m i s + 1 s + 1   f ( i ) = ∑ i s = m i s + 1 [ ∑ i = m i s s f ( i ) ] = ∑ i s = m i s + 1 [ ∑ i = m i s ( i s − i + s − 1 s − 1 ) f ( i ) ] = ∑ i = m i s + 1 ( i s + 1 − i + ( s + 1 ) − 1 ( s + 1 ) − 1 ) f ( i ) {\displaystyle {\begin{aligned}p\geq 1,\quad &\sum _{i=m}^{i_{p}}\!{}^{p}\ f(i)&=&\sum _{i=m}^{i_{p}}{i_{p}-i+p-1 \choose p-1}f(i)\\p=1,\quad &\sum _{i=m}^{i_{1}}\!{}^{1}\ f(i)&=&\sum _{i=m}^{i_{1}}{i_{1}-i \choose 0}f(i)=\sum _{i=m}^{i_{1}}f(i)\\p=s,\quad &\sum _{i=m}^{i_{s}}\!{}^{s}\ f(i)&=&\sum _{i=m}^{i_{s}}{i_{s}-i+s-1 \choose s-1}f(i)\\p=s+1,\quad &\sum _{i=m}^{i_{s+1}}\!{}^{s+1}\ f(i)&=&\sum _{i_{s}=m}^{i_{s+1}}\left[\sum _{i=m}^{i_{s}}\!{}^{s},円{\mathit {f}}(i)\right]\\&&=&\sum _{i_{s}=m}^{i_{s+1}}\left[\sum _{i=m}^{i_{s}}{i_{s}-i+s-1 \choose s-1}f(i)\right]\\&&=&\sum _{i=m}^{i_{s+1}}{i_{s+1}-i+(s+1)-1 \choose (s+1)-1}f(i)\\\end{aligned}},円} {\displaystyle {\begin{aligned}p\geq 1,\quad &\sum _{i=m}^{i_{p}}\!{}^{p}\ f(i)&=&\sum _{i=m}^{i_{p}}{i_{p}-i+p-1 \choose p-1}f(i)\\p=1,\quad &\sum _{i=m}^{i_{1}}\!{}^{1}\ f(i)&=&\sum _{i=m}^{i_{1}}{i_{1}-i \choose 0}f(i)=\sum _{i=m}^{i_{1}}f(i)\\p=s,\quad &\sum _{i=m}^{i_{s}}\!{}^{s}\ f(i)&=&\sum _{i=m}^{i_{s}}{i_{s}-i+s-1 \choose s-1}f(i)\\p=s+1,\quad &\sum _{i=m}^{i_{s+1}}\!{}^{s+1}\ f(i)&=&\sum _{i_{s}=m}^{i_{s+1}}\left[\sum _{i=m}^{i_{s}}\!{}^{s},円{\mathit {f}}(i)\right]\\&&=&\sum _{i_{s}=m}^{i_{s+1}}\left[\sum _{i=m}^{i_{s}}{i_{s}-i+s-1 \choose s-1}f(i)\right]\\&&=&\sum _{i=m}^{i_{s+1}}{i_{s+1}-i+(s+1)-1 \choose (s+1)-1}f(i)\\\end{aligned}},円}

Miscellaneous items (some valid, some not)

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= ∑ i = m i s + 1 f ( i ) ∑ j = 0 i s + 1 − i ( j + s − 1 s − 1 ) = ∑ i = m i s + 1 f ( i ) ∑ j = s − 1 i s + 1 − i + s − 1 ( [ j − ( s − 1 ) ] + s − 1 s − 1 ) = ∑ i = m i s + 1 f ( i ) ∑ j = s − 1 i s + 1 − i + s − 1 ( j s − 1 ) = ∑ i = m i s + 1 ( i s + 1 − i + ( s + 1 ) − 1 ( s + 1 ) − 1 ) f ( i ) {\displaystyle {\begin{aligned}&=\sum _{i=m}^{i_{s+1}}f(i)\sum _{j=0}^{i_{s+1}-i}{j+s-1 \choose s-1}\\&=\sum _{i=m}^{i_{s+1}}f(i)\sum _{j=s-1}^{i_{s+1}-i+s-1}{[j-(s-1)]+s-1 \choose s-1}\\&=\sum _{i=m}^{i_{s+1}}f(i)\sum _{j=s-1}^{i_{s+1}-i+s-1}{j \choose s-1}\\&=\sum _{i=m}^{i_{s+1}}{i_{s+1}-i+(s+1)-1 \choose (s+1)-1}f(i)\end{aligned}},円} {\displaystyle {\begin{aligned}&=\sum _{i=m}^{i_{s+1}}f(i)\sum _{j=0}^{i_{s+1}-i}{j+s-1 \choose s-1}\\&=\sum _{i=m}^{i_{s+1}}f(i)\sum _{j=s-1}^{i_{s+1}-i+s-1}{[j-(s-1)]+s-1 \choose s-1}\\&=\sum _{i=m}^{i_{s+1}}f(i)\sum _{j=s-1}^{i_{s+1}-i+s-1}{j \choose s-1}\\&=\sum _{i=m}^{i_{s+1}}{i_{s+1}-i+(s+1)-1 \choose (s+1)-1}f(i)\end{aligned}},円}



= ∑ i = m i s + 1 ( i s + 1 − i + s − 1 s − 1 ) ∑ j = m i f ( j ) {\displaystyle {\begin{aligned}&=\sum _{i=m}^{i_{s+1}}{i_{s+1}-i+s-1 \choose s-1}\sum _{j=m}^{i}f(j)\end{aligned}},円} {\displaystyle {\begin{aligned}&=\sum _{i=m}^{i_{s+1}}{i_{s+1}-i+s-1 \choose s-1}\sum _{j=m}^{i}f(j)\end{aligned}},円}



f ( x ) = n − 0 h − 0 x + 0 = n h x {\displaystyle f(x)={\dfrac {n-0}{h-0}}x+0={\dfrac {n}{h}}x,円} {\displaystyle f(x)={\dfrac {n-0}{h-0}}x+0={\dfrac {n}{h}}x,円}

F ( x ) = n 2 h x 2 + C {\displaystyle F(x)={\dfrac {n}{2h}}x^{2}+C,円} {\displaystyle F(x)={\dfrac {n}{2h}}x^{2}+C,円}

g ( x ) = n − b h − 0 x + b = n − b h x + b {\displaystyle g(x)={\dfrac {n-b}{h-0}}x+b={\dfrac {n-b}{h}}x+b,円} {\displaystyle g(x)={\dfrac {n-b}{h-0}}x+b={\dfrac {n-b}{h}}x+b,円}

G ( x ) = n − b 2 h x 2 + b x + C {\displaystyle G(x)={\dfrac {n-b}{2h}}x^{2}+bx+C,円} {\displaystyle G(x)={\dfrac {n-b}{2h}}x^{2}+bx+C,円}

A f = F ( h ) − F ( 0 ) = [ n 2 h ( h ) 2 + C ] − [ n 2 h ( 0 ) 2 + C ] = 1 2 n h {\displaystyle A_{f}=F(h)-F(0)=\left[{\dfrac {n}{2h}}(h)^{2}+C\right]-\left[{\dfrac {n}{2h}}(0)^{2}+C\right]={\tfrac {1}{2}}nh,円} {\displaystyle A_{f}=F(h)-F(0)=\left[{\dfrac {n}{2h}}(h)^{2}+C\right]-\left[{\dfrac {n}{2h}}(0)^{2}+C\right]={\tfrac {1}{2}}nh,円}

A g = G ( h ) − G ( 0 ) = [ n − b 2 h ( h ) 2 + b ( h ) + C ] − [ n − b 2 h ( 0 ) 2 + b ( 0 ) + C ] = 1 2 n h − 1 2 b h + b h = 1 2 n h + 1 2 b h {\displaystyle A_{g}=G(h)-G(0)=\left[{\dfrac {n-b}{2h}}(h)^{2}+b(h)+C\right]-\left[{\frac {n-b}{2h}}(0)^{2}+b(0)+C\right]={\tfrac {1}{2}}nh-{\tfrac {1}{2}}bh+bh={\tfrac {1}{2}}nh+{\tfrac {1}{2}}bh,円} {\displaystyle A_{g}=G(h)-G(0)=\left[{\dfrac {n-b}{2h}}(h)^{2}+b(h)+C\right]-\left[{\frac {n-b}{2h}}(0)^{2}+b(0)+C\right]={\tfrac {1}{2}}nh-{\tfrac {1}{2}}bh+bh={\tfrac {1}{2}}nh+{\tfrac {1}{2}}bh,円}

A = A g − A f = ( 1 2 n h + 1 2 b h ) − ( 1 2 n h ) = 1 2 b h {\displaystyle A=A_{g}-A_{f}=\left({\tfrac {1}{2}}nh+{\tfrac {1}{2}}bh\right)-\left({\tfrac {1}{2}}nh\right)=\mathbf {{\tfrac {1}{2}}bh} ,円} {\displaystyle A=A_{g}-A_{f}=\left({\tfrac {1}{2}}nh+{\tfrac {1}{2}}bh\right)-\left({\tfrac {1}{2}}nh\right)=\mathbf {{\tfrac {1}{2}}bh} ,円}


= ( i k + 1 − m + 1 ) ( k − 1 k − 1 ) ∑ i = m i k + 1 f ( i ) + ( i k + 1 − m ) ( k k − 1 ) ∑ i = m i k + 1 − 1 f ( i ) + ( i k + 1 − m − 1 ) ( k + 1 k − 1 ) ∑ i = m i k + 1 − 2 f ( i ) + … + 3 ( i k + 1 − m + 2 + k − 1 k − 1 ) ∑ i = m m + 2 f ( i ) + 2 ( i k + 1 − m + 1 + k − 1 k − 1 ) ∑ i = m m + 1 f ( i ) + ( i k + 1 − m + k − 1 k − 1 ) ∑ i = m m f ( i ) {\displaystyle {\begin{aligned}&=(i_{k+1}-m+1){k-1 \choose k-1}\sum _{i=m}^{i_{k+1}}f(i)+(i_{k+1}-m){k \choose k-1}\sum _{i=m}^{i_{k+1}-1}f(i)\\&+(i_{k+1}-m-1){k+1 \choose k-1}\sum _{i=m}^{i_{k+1}-2}f(i)+\ldots +3{i_{k+1}-m+2+k-1 \choose k-1}\sum _{i=m}^{m+2}f(i)\\&+2{i_{k+1}-m+1+k-1 \choose k-1}\sum _{i=m}^{m+1}f(i)+{i_{k+1}-m+k-1 \choose k-1}\sum _{i=m}^{m}f(i)\end{aligned}},円} {\displaystyle {\begin{aligned}&=(i_{k+1}-m+1){k-1 \choose k-1}\sum _{i=m}^{i_{k+1}}f(i)+(i_{k+1}-m){k \choose k-1}\sum _{i=m}^{i_{k+1}-1}f(i)\\&+(i_{k+1}-m-1){k+1 \choose k-1}\sum _{i=m}^{i_{k+1}-2}f(i)+\ldots +3{i_{k+1}-m+2+k-1 \choose k-1}\sum _{i=m}^{m+2}f(i)\\&+2{i_{k+1}-m+1+k-1 \choose k-1}\sum _{i=m}^{m+1}f(i)+{i_{k+1}-m+k-1 \choose k-1}\sum _{i=m}^{m}f(i)\end{aligned}},円}

これ、ちょっとちがうね。


= f ( m ) ∑ j = m i k + 1 ( i k + 1 − j + 1 ) ( j − m + k − 1 k − 1 ) + f ( m + 1 ) ∑ j = m i k + 1 − 1 ( i k + 1 − j + 1 ) ( j − m + k − 1 k − 1 ) + f ( m + 2 ) ∑ j = m i k + 1 − 2 ( i k + 1 − j + 1 ) ( j − m + k − 1 k − 1 ) + … + f ( i k + 1 − 2 ) ∑ j = m m + 2 ( i k + 1 − j + 1 ) ( j − m + k − 1 k − 1 ) + f ( i k + 1 − 1 ) ∑ j = m m + 1 ( i k + 1 − j + 1 ) ( j − m + k − 1 k − 1 ) + f ( i k + 1 ) ∑ j = m m ( i k + 1 − j + 1 ) ( j − m + k − 1 k − 1 ) {\displaystyle {\begin{aligned}&=f(m)\sum _{j=m}^{i_{k+1}}(i_{k+1}-j+1){j-m+k-1 \choose k-1}+f(m+1)\sum _{j=m}^{i_{k+1}-1}(i_{k+1}-j+1){j-m+k-1 \choose k-1}\\&+f(m+2)\sum _{j=m}^{i_{k+1}-2}(i_{k+1}-j+1){j-m+k-1 \choose k-1}+\ldots +f(i_{k+1}-2)\sum _{j=m}^{m+2}(i_{k+1}-j+1){j-m+k-1 \choose k-1}\\&+f(i_{k+1}-1)\sum _{j=m}^{m+1}(i_{k+1}-j+1){j-m+k-1 \choose k-1}+f(i_{k+1})\sum _{j=m}^{m}(i_{k+1}-j+1){j-m+k-1 \choose k-1}\end{aligned}},円} {\displaystyle {\begin{aligned}&=f(m)\sum _{j=m}^{i_{k+1}}(i_{k+1}-j+1){j-m+k-1 \choose k-1}+f(m+1)\sum _{j=m}^{i_{k+1}-1}(i_{k+1}-j+1){j-m+k-1 \choose k-1}\\&+f(m+2)\sum _{j=m}^{i_{k+1}-2}(i_{k+1}-j+1){j-m+k-1 \choose k-1}+\ldots +f(i_{k+1}-2)\sum _{j=m}^{m+2}(i_{k+1}-j+1){j-m+k-1 \choose k-1}\\&+f(i_{k+1}-1)\sum _{j=m}^{m+1}(i_{k+1}-j+1){j-m+k-1 \choose k-1}+f(i_{k+1})\sum _{j=m}^{m}(i_{k+1}-j+1){j-m+k-1 \choose k-1}\end{aligned}},円}

= ∑ i = m i k + 1 f ( i ) ∑ j = 0 i k + 1 − i ( i k + 1 − j − m + 1 ) ( j + k − 1 k − 1 ) {\displaystyle =\sum _{i=m}^{i_{k+1}}f(i)\sum _{j=0}^{i_{k+1}-i}(i_{k+1}-j-m+1){j+k-1 \choose k-1},円} {\displaystyle =\sum _{i=m}^{i_{k+1}}f(i)\sum _{j=0}^{i_{k+1}-i}(i_{k+1}-j-m+1){j+k-1 \choose k-1},円} これもちがう。

AltStyle によって変換されたページ (->オリジナル) /