User:Cornince

Alternative account: User:Beneficii

Basic definition of a sum
$$\sum_{i=m}^{n} \mathit{f}(i) = \mathit{f}(m) + \mathit{f}(m+1) + \mathit{f}(m+2) + \, ... \, + \mathit{f}(n-2) + \mathit{f}(n-1) + \mathit{f}(n) \,$$

Recursive summation
Where $$p \ge 0$$:

$$\sum_{i=m}^{i_p} \! {}^p \ f(i) = \sum_{i_{p-1}=m}^{i_p} \, \sum_{i_{p-2}=m}^{i_{p-1}} \, \sum_{i_{p-3}=m}^{i_{p-2}} ... \sum_{i_2=m}^{i_3} \, \sum_{i_1=m}^{i_2} \, \sum_{i_0=m}^{i_1} f(i_0) \,$$

$$\sum_{i=m}^{i_0} \! {}^0 \ \mathit{f}(i) = \mathit{f}(i_0)\,$$

$$\sum_{i=m}^{i_1} \! {}^1 \ \mathit{f}(i) = \sum_{i_0=m}^{i_1} \mathit{f}(i_0)\,$$

$$\mathrm{If \ } \mathit{g}(n) = \sum_{i=m}^n \! {}^{-p} \, \mathit{f}(i), \ \mathrm{then \ } \mathit{f}(n) = \sum_{i=m}^n \! {}^{p} \, \mathit{g}(i). \,$$

Where $$p > 0$$:

$$\sum_{i=m}^{i_p} \! {}^p \ \mathit{f}(i) = \sum_{i_{p-1}=m}^{i_p} \left [ \sum_{i=m}^{i_{p-1}} \! {}^{p-1} \, \mathit{f}(i) \right ] \,$$

$$ \sum_{i_{p-1}=m}^{i_p} \left [ \sum_{i=m}^{i_{p-1}} \! {}^{p-1} \, \mathit{f}(i) \right ] = \sum_{i=m}^{m} \! {}^{p-1} \, \mathit{f}(i) + \sum_{i=m}^{m+1} \! {}^{p-1} \, \mathit{f}(i) + \sum_{i=m}^{m+2} \! {}^{p-1} \, \mathit{f}(i) + \, ... + \, \sum_{i=m}^{i_p - 2} \! {}^{p-1} \, \mathit{f}(i) + \sum_{i=m}^{i_p - 1} \! {}^{p-1} \, \mathit{f}(i) + \sum_{i=m}^{i_p} \! {}^{p-1} \, \mathit{f}(i) \,$$

Shifting of starting and ending indices
$$\sum_{i=m}^{n} \mathit{f}(i) = \sum_{i=m+u}^{n+u} \mathit{f}(i - u) \,$$

Proof of the equality of the shifting of indices:
$$\,\! \begin{array}{lcl} \sum_{i=m+u}^{n+u} \mathit{f}(i - u) & = & \mathit{f}((m+u) - u) + \mathit{f}((m+u+1) - u) + \mathit{f}((m+u+2) - u) \\ & & + \, ... \, + \mathit{f}((n+u-2) - u) + \mathit{f}((n+u-1) - u) + \mathit{f}((n+u) - u)       \\ \\ & = & \mathit{f}(m) + \mathit{f}(m+1) + \mathit{f}(m+2) + \, ... \, + \mathit{f}(n-2) + \mathit{f}(n-1) + \mathit{f}(n) \\ \\ & = & \sum_{i=m}^{n} \mathit{f}(i) \end{array} \,$$

Smaller summation notation
$$\,\! \sum_{i=m}^{n} \! {}^p \ \mathit{f}(i) = \textstyle \! {}_p \! \sum_{i=m}^n \mathit{f}(i)\,$$

Recursive geometric series
$$\,\! \begin{array}{lcl} {}_p \! \sum_{i=m}^n r^i & = & {r^{n+p} \over (r-1)^p} - \sum_{k=0}^{p-1} {r^{m + p - (k+1)} \prod_{j=1}^k (n - m + j) \over k! (r-1)^{p-k}}     \\ & = & {r^{n+p} \over (r-1)^p} - \sum_{k=0}^{p-1} {r^{m + p - (k+1)} {(n - m + k)! \over (n - m)!} \over (r-1)^{p-k} k!} \\ & = & {r^{n+p} \over (r-1)^p} - \sum_{k=0}^{p-1} \left ({r^{m + p - (k+1)} \over (r-1)^{p-k}} \right ) {{n - m + k} \choose k} \end{array}\,$$

Combinations proof (used in below proof)
$$\mathrm{For\ non-negative\ integers\ } q\ \mathrm{and\ } t \ge q, \quad \sum_{z=q}^{t} {z \choose q} = {t + 1 \choose q + 1}. \,$$

$$t = q, \quad \sum_{z=q}^{q} {z \choose q} = {q \choose q} = 1 = {q + 1 \choose q + 1} \,$$

$$t = c, \quad \sum_{z=q}^{c} {z \choose q} = {c + 1 \choose q + 1} \,$$

$$t = c + 1, \quad \sum_{z=q}^{c+1} {z \choose q} = \sum_{z=q}^{c} {z \choose q} + {c + 1 \choose q} = {c + 1 \choose q + 1} + {c + 1 \choose q}  = {(c + 1) + 1 \choose q + 1} \,$$

Definition
$$\mathrm{For \ a\ real\ number\ } r \ne 1\ \mathrm{and\ non-negative\ integers\ } m,\ p,\ \mathrm{and\ } i_p,\ \mathrm{where\ } i_p \ge m, $$

$$ \sum_{i=m}^{i_p} \!\! {}^p \, r^i = {r^{i_p+p} \over (r-1)^p} - \sum_{k=0}^{p-1} {r^{m + p - (k+1)} \prod_{j=1}^k (i_p - m + j) \over k! (r-1)^{p-k}}. \,$$

Base case (and some specific examples)
$$p = 0, \quad \sum_{i=m}^{i_0} \!\! {}^0 \, r^i = \cfrac{r^{i_0+0}}{(r-1)^0} = r^{i_0} \,$$

$$p = 1, \quad \sum_{i=m}^{i_1} \!\! {}^1 \, r^i = \cfrac{r^{i_1+1}}{(r-1)^1} - \cfrac{r^m}{0! (r-1)^1} = \cfrac{r^{i_1+1} - r^{m}}{r-1} \,$$

$$p = 2, \quad \sum_{i=m}^{i_2} \!\! {}^2 \, r^i = \cfrac{r^{i_2+2}}{(r-1)^2} - \cfrac{r^{m+1}}{0! (r-1)^2} - \cfrac{r^m (i_2 - m + 1)}{1! (r-1)^1} = \cfrac{\cfrac{r^{i_2+2} - r^{m+1}}{r-1} - r^m (i_2 - m + 1)}{r-1} \,$$

$$p = 3, \,$$

$$\quad \sum_{i=m}^{i_3} \!\! {}^3 \, r^i = \cfrac{r^{i_3+3}}{(r-1)^3} - \cfrac{r^{m+2}}{0! (r-1)^3} - \cfrac{r^{m+1} (i_3 - m + 1)}{1! (r-1)^2} - \cfrac{r^m (i_3 - m + 1)(i_3 - m + 2)}{2! (r-1)^1} \,$$



\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} \,$$

$$p = 4, \,$$

$$\quad \sum_{i=m}^{i_4} \!\! {}^4 \, r^i = \cfrac{r^{i_4+4}}{(r-1)^4} - \cfrac{r^{m+3}}{0! (r-1)^4} - \cfrac{r^{m+2} (i_4 - m + 1)}{1! (r-1)^3} - \cfrac{r^{m+1} (i_4 - m + 1)(i_4 - m + 2)}{2! (r-1)^2} \,$$



- \cfrac{r^{m} (i_4 - m + 1)(i_4 - m + 2)(i_4 - m + 3)}{3! (r-1)^1} \,$$



\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} \,$$

Inductive step
$$p = a, \quad \sum_{i=m}^{i_a} \!\! {}^a \, r^i = {r^{{i_a}+a} \over (r-1)^a} - \sum_{k=0}^{a-1} {r^{m + a - (k+1)} \prod_{j=1}^k (i_a - m + j) \over k! (r-1)^{a-k}} \,$$

$$p = a+1, \quad \sum_{i=m}^{i_{(a+1)}} \!\! {}^{(a+1)} \, r^i = \sum_{{i_a}=m}^{i_{(a+1)}} \left [ \sum_{i=m}^{i_a} \!\! {}^a \, r^i \right ] \,$$

$$ = \sum_{{i_a}=m}^{i_{(a+1)}} \left [ {r^{{i_a}+a} \over (r-1)^a} - \sum_{k=0}^{a-1} {r^{m + a - (k+1)} \prod_{j=1}^k (i_a - m + j) \over k! (r-1)^{a-k}} \right ] \,$$

$$ = \sum_{{i_a}=m}^{i_{(a+1)}} \left [ {r^{{i_a}+a} \over (r-1)^a} \right ] - \sum_{{i_a}=m}^{i_{(a+1)}} \left [ \sum_{k=0}^{a-1} {r^{m + a - (k+1)} \prod_{j=1}^k (i_a - m + j) \over k! (r-1)^{a-k}} \right ] \,$$

$$ = {\sum_{{i_a}=m}^{i_{(a+1)}} \left ( r^{{i_a}+a} \right ) \over (r-1)^a} - \sum_{{i_a}=m}^{i_{(a+1)}} \left [ \sum_{k=0}^{a-1} {r^{m + a - (k+1)} \left [{(i_a - m + k)! \over (i_a - m)!} \right ] \over (r-1)^{a-k} \qquad k!} \right ] \,$$

$$ = {r^a \sum_{{i_a}=m}^{i_{(a+1)}} \left ( r^{i_a} \right ) \over (r-1)^a} - \sum_{{i_a}=m}^{i_{(a+1)}} \left [ \sum_{k=0}^{a-1} \left ( {r^{m + a - (k+1)} \over (r-1)^{a-k}} \right ) {i_a - m + k \choose k} \right ] \,$$

$$ = {r^a \left ( {r^{i_{(a+1)} + 1} - r^m \over r-1} \right ) \over (r-1)^a} - \sum_{k=0}^{a-1} \left [ \left ( {r^{m + a - (k+1)} \over (r-1)^{a-k}} \right ) \sum_{{i_a}=m}^{i_{(a+1)}} {i_a - m + k \choose k} \right ] \,$$

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

$$= {r^{i_{(a+1)} + a + 1} - r^{m+a} \over (r-1)^{a+1}} - \sum_{k=0}^{a-1} \left [ \left ( {r^{m + a - (k+1)} \over (r-1)^{a-k}} \right ) \sum_{i_a=k}^{i_{(a+1)} - m + k} {i_a \choose k} \right ] \,$$

See combinations proof above:

$$ = {r^{i_{(a+1)} + a + 1} \over (r-1)^{a+1}} - {r^{m+a} \over (r-1)^{a+1}} - \sum_{k=0}^{a-1} \left ( {r^{m + a - (k+1)} \over (r-1)^{a-k}} \right ) {i_{(a+1)} - m + k + 1 \choose k + 1} \,$$

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

$$ = {r^{i_{(a+1)} + a + 1} \over (r-1)^{a+1}}  - {r^{m+a} \over (r-1)^{a+1}} - \sum_{k=1}^{(a+1)-1} \left ( {r^{m + a - ((k - 1)+1)} \over (r-1)^{a-(k - 1)}} \right ) {i_{(a+1)} - m + (k - 1) + 1 \choose (k - 1) + 1} \,$$

$$ = {r^{i_{(a+1)} + a + 1} \over (r-1)^{a+1}} - {r^{m+a} \over (r-1)^{a+1}} - \sum_{k=1}^{(a+1)-1} \left ( {r^{m + a - k + 1 - 1} \over (r-1)^{a - k + 1}} \right ) {i_{(a+1)} - m + k \choose k} \,$$

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

$$ = {r^{i_{(a+1)} + (a + 1)} \over (r-1)^{(a+1)}} - {r^{m+a} \over (r-1)^{a+1}} - \left [ \sum_{k=0}^{(a+1)-1} \left ( {r^{m + (a + 1) - k - 1} \over (r-1)^{(a + 1) - k}} \right ) {i_{(a+1)} - m + k \choose k} - \left ( {r^{m+a} \over (r-1)^{a+1}} \right ) {i_{(a+1)} - m \choose 0} \right ] \,$$

Terms cancel out.

$$ = {r^{i_{(a+1)} + (a + 1)} \over (r-1)^{(a+1)}} - \sum_{k=0}^{(a+1)-1} \left ( {r^{m + (a + 1) - (k + 1)} \over (r-1)^{(a + 1) - k}} \right ) {i_{(a+1)} - m + k \choose k} \,$$

$$ = {r^{i_{(a+1)} + (a + 1)} \over (r-1)^{(a+1)}} - \sum_{k=0}^{(a+1)-1} {r^{m + (a + 1) - (k+1)} \left [{(i_{(a+1)} - m + k)! \over (i_{(a+1)} - m)!} \right ] \over (r-1)^{(a+1)-k} \qquad k!} \,$$

$$\sum_{i=m}^{i_{(a+1)}} \!\! {}^{(a+1)} \, r^i = {r^{i_{(a+1)} + (a + 1)} \over (r-1)^{(a+1)}} - \sum_{k=0}^{(a+1)-1} {r^{m + (a+1) - (k+1)} \prod_{j=1}^k (i_{(a+1)} - m + j) \over k! (r-1)^{(a+1)-k}} \,$$

Q.E.D.

One method
$$\begin{align} n \ge 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{align}\,$$

Inductive method
$$\begin{align} n \ge 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{align}\,$$

Second proof, this one for the general formula for recursive summation series
$$\begin{align} p \ge 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 \, \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{align}\,$$

Miscellaneous items (some valid, some not)
$$\begin{align} &= \sum_{i=m}^{i_{s+1}} f(i) \sum_{j=0}^{i_{s+1}-i} {j + s - 1 \choose s - 1} \\ &= \sum_{i=m}^{i_{s+1}} f(i) \sum_{j=s-1}^{i_{s+1}-i+s-1} {[j - (s-1)] + s - 1 \choose s - 1} \\ &= \sum_{i=m}^{i_{s+1}} f(i) \sum_{j=s-1}^{i_{s+1}-i+s-1} {j \choose s - 1} \\ &= \sum_{i=m}^{i_{s+1}} {i_{s+1} - i + (s+1) - 1 \choose (s+1) - 1} f(i) \end{align}\,$$

$$\begin{align} &= \sum_{i=m}^{i_{s+1}} {i_{s+1} - i + s - 1 \choose s - 1} \sum_{j=m}^{i} f(j) \end{align}\,$$

$$f(x) = \dfrac{n-0}{h-0}x + 0 = \dfrac{n}{h}x \,$$

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

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

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

$$A_f = F(h) - F(0) = \left [ \dfrac{n}{2h}(h)^2 + C \right ] - \left [ \dfrac{n}{2h}(0)^2 + C \right ] = \tfrac{1}{2} nh \,$$

$$A_g = G(h) - G(0) = \left [ \dfrac{n-b}{2h}(h)^2 + b(h) + C \right ] - \left [ \frac{n-b}{2h}(0)^2 + b(0) + C \right ] = \tfrac{1}{2}nh - \tfrac{1}{2}bh + bh = \tfrac{1}{2}nh + \tfrac{1}{2}bh \,$$

$$A = A_g - A_f = \left ( \tfrac{1}{2}nh + \tfrac{1}{2}bh \right ) - \left ( \tfrac{1}{2} nh \right ) = \mathbf{\tfrac{1}{2}bh} \,$$

$$\begin{align} &= (i_{k+1} - m + 1) {k-1 \choose k-1} \sum_{i=m}^{i_{k+1}} f(i) + (i_{k+1} - m) {k \choose k-1} \sum_{i=m}^{i_{k+1}-1} f(i) \\ &+ (i_{k+1} - m - 1) {k+1 \choose k-1} \sum_{i=m}^{i_{k+1}-2} f(i) + \ldots + 3 {i_{k+1} - m + 2 + k-1 \choose k-1} \sum_{i=m}^{m+2} f(i) \\ &+ 2 {i_{k+1} - m + 1 + k-1 \choose k-1} \sum_{i=m}^{m+1} f(i) + {i_{k+1} - m + k-1 \choose k-1} \sum_{i=m}^{m} f(i) \end{align}\,$$

これ、ちょっとちがうね.

$$\begin{align} &= f(m) \sum_{j=m}^{i_{k+1}} (i_{k+1} - j + 1) {j - m + k - 1 \choose k - 1} + f(m+1) \sum_{j=m}^{i_{k+1} - 1} (i_{k+1} - j + 1) {j - m + k - 1 \choose k - 1} \\ &+ f(m+2) \sum_{j=m}^{i_{k+1} - 2} (i_{k+1} - j + 1) {j - m + k - 1 \choose k - 1} + \ldots + f(i_{k+1}-2) \sum_{j=m}^{m+2} (i_{k+1} - j + 1) {j - m + k - 1 \choose k - 1} \\ &+ f(i_{k+1}-1) \sum_{j=m}^{m+1} (i_{k+1} - j + 1) {j - m + k - 1 \choose k - 1} + f(i_{k+1}) \sum_{j=m}^{m} (i_{k+1} - j + 1) {j - m + k - 1 \choose k - 1} \end{align}\,$$

$$=\sum_{i=m}^{i_{k+1}} f(i) \sum_{j=0}^{i_{k+1} - i} (i_{k+1} - j - m + 1) {j + k - 1 \choose k - 1} \,$$ これもちがう.