積分法｜区分求積法を扱った入試問題を解いてみよう

\begin{align*} &\sum_{k=1}^{2n} (-1)^{\scriptsize{k}} \ \left(\frac{k}{2n} \right)^{\scriptsize{100}} \\[ 10pt ] = &\sum_{l=1}^{n} \cfrac{(-1)^{\scriptsize{2l}} \cdot (2l)^{\scriptsize{100}} + (-1)^{\scriptsize{2l-1}} \cdot (2l-1)^{\scriptsize{100}}}{(2n)^{\scriptsize{100}}} \\[ 10pt ] = &\sum_{l=1}^{n} \cfrac{(2l)^{\scriptsize{100}} \ – (2l-1)^{\scriptsize{100}}}{(2n)^{\scriptsize{100}}} \\[ 10pt ] = &\sum_{l=1}^{n} \frac{1}{(2n)^{\scriptsize{100}}} \left[ (2l)^{\scriptsize{100}} \ – \left\{ \sum_{p=0}^{98} {}_{100} \mathrm{ C }_{p} \ (-1)^{\scriptsize{100-p}} \ (2l)^{\scriptsize{p}} -100 \cdot (2l)^{\scriptsize{99}} + (2l)^{\scriptsize{100}} \right\} \right] \\[ 10pt ] = &\sum_{l=1}^{n} \frac{1}{(2n)^{\scriptsize{100}}} \left\{ 100 \cdot (2l)^{\scriptsize{99}} \ – \sum_{p=0}^{98} {}_{100} \mathrm{ C }_{p} \ (-1)^{\scriptsize{100-p}} \ (2l)^{\scriptsize{p}} \right\} \\[ 10pt ] = &\sum_{l=1}^{n} \left\{ 100 \cdot \frac{(2l)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \ – \frac{1}{(2n)^{\scriptsize{100}}} \sum_{p=0}^{98} {}_{100} \mathrm{ C }_{p} \ (-1)^{\scriptsize{100-p}} \ (2l)^{\scriptsize{p}} \right\} \\[ 10pt ] = &\sum_{l=1}^{n} \frac{50}{n} \left(\frac{l}{n} \right)^{\scriptsize{99}} \ – \sum_{l=1}^{n} \left\{ \sum_{p=0}^{98} {}_{100} \mathrm{ C }_{p} \ (-1)^{\scriptsize{100-p}} \cdot \cfrac{1}{2^{\scriptsize{100-p}}} \cdot \cfrac{1}{n^{\scriptsize{100-p}}} \cdot \left(\frac{l}{n} \right)^{\scriptsize{p}} \right\} \\[ 10pt ] = &\frac{50}{n} \sum_{l=1}^{n} \left(\frac{i}{n} \right)^{\scriptsize{99}} \ – \sum_{p=0}^{98} {}_{100} \mathrm{ C }_{p} \ (-1)^{\scriptsize{100-p}} \cdot \cfrac{1}{2^{\scriptsize{100-p}}} \cdot \cfrac{1}{n^{\scriptsize{99-p}}} \cdot \frac{1}{n} \sum_{l=1}^{n} \left(\frac{l}{n} \right)^{\scriptsize{p}} \end{align*}

$n \to \infty$ のとき

\begin{align*} &\int_{0}^{1} {50x^{\scriptsize{99}} dx} \ – \sum_{p=0}^{98} {}_{100} \mathrm{ C }_{p} \ (-1)^{\scriptsize{100-p}} \cdot \cfrac{1}{2^{\scriptsize{100-p}}} \cdot 0 \cdot \int_{0}^{1} {x^{\scriptsize{p}} dx} \\[ 10pt ] = \ &\int_{0}^{1} {50x^{\scriptsize{99}} dx} \\[ 10pt ] = \ &\left[ \ \frac{1}{2} x^{\scriptsize{100}} \ \right]_0^1 \\[ 10pt ] = \ &\frac{1}{2} \end{align*} \begin{align*} &\text{したがって} \\[ 5pt ] &\quad \lim_{n \to \infty} \sum_{k=1}^{2n} (-1)^{\scriptsize{k}} \ \left(\frac{k}{2n} \right)^{\scriptsize{100}} = \frac{1}{2} \end{align*}

\begin{align*} &\quad \sum_{k=1}^{2n} (-1)^{\scriptsize{k}} \ \left(\frac{k}{2n} \right)^{\scriptsize{100}} = \sum_{l=1}^{n} \cfrac{(2l)^{\scriptsize{100}} \ – (2l-1)^{\scriptsize{100}}}{(2n)^{\scriptsize{100}}} \\[ 10pt ] &\text{$f(x) = x^{\scriptsize{100}}$ とおくと} \\[ 5pt ] &\quad f'(x) = 100x^{\scriptsize{99}} \\[ 10pt ] &\text{平均値の定理から} \\[ 5pt ] &\quad f(2l)-f(2l-1) = \left\{ (2l)-(2l-1) \right\} \cdot 100c^{\scriptsize{99}} \\[ 10pt ] &\qquad 2l-1 \lt c \lt 2l \\[ 5pt ] &\text{を満たす $c$ が存在する。} \\[ 5pt ] &\quad (2l)^{\scriptsize{100}}-(2l-1)^{\scriptsize{100}} = 100c^{\scriptsize{99}} \\[ 10pt ] &\text{より} \\[ 5pt ] &\quad 100(2l-1)^{\scriptsize{99}} \lt (2l)^{\scriptsize{100}}-(2l-1)^{\scriptsize{100}} \lt 100(2l)^{\scriptsize{99}} \\[ 10pt ] &\text{よって} \\[ 5pt ] &\quad 100 \sum_{l=1}^{n} \frac{(2l-1)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \lt \sum_{l=1}^{n} \cfrac{(2l)^{\scriptsize{100}} \ – (2l-1)^{\scriptsize{100}}}{(2n)^{\scriptsize{100}}} \lt 100 \sum_{l=1}^{n} \frac{(2l)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \\[ 10pt ] &\quad 100 \sum_{l=1}^{n} \frac{(2l-1)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \lt \sum_{k=1}^{2n} (-1)^{\scriptsize{k}} \ \left(\frac{k}{2n} \right)^{\scriptsize{100}} \lt 100 \sum_{l=1}^{n} \frac{(2l)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \\[ 10pt ] &\text{ここで} \\[ 5pt ] &\quad 100 \sum_{l=1}^{n} \frac{(2l-2)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \lt 100 \sum_{l=1}^{n} \frac{(2l-1)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \\[ 5pt ] &\text{より} \\[ 5pt ] &\quad 100 \sum_{l=1}^{n} \frac{(2l-2)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \lt \sum_{k=1}^{2n} (-1)^{\scriptsize{k}} \ \left(\frac{k}{2n} \right)^{\scriptsize{100}} \lt 100 \sum_{l=1}^{n} \frac{(2l)^{\scriptsize{99}}}{(2n)^{\scriptsize{100}}} \\[ 10pt ] &\quad \therefore \frac{50}{n} \sum_{l=1}^{n} \left(\frac{l-1}{n} \right)^{\scriptsize{99}} \lt \sum_{k=1}^{2n} (-1)^{\scriptsize{k}} \ \left(\frac{k}{2n} \right)^{\scriptsize{100}} \lt \frac{50}{n} \sum_{l=1}^{n} \left(\frac{l}{n} \right)^{\scriptsize{99}} \\[ 10pt ] &\text{$n \to \infty$ のとき} \\[ 5pt ] &\quad \frac{50}{n} \sum_{l=1}^{n} \left(\frac{l-1}{n} \right)^{\scriptsize{99}} \longrightarrow \ \int_{0}^{1} {50x^{\scriptsize{99}} dx} = \left[ \ \frac{1}{2} x^{\scriptsize{100}} \ \right]_0^1 = \frac{1}{2} \\[ 10pt ] &\quad \frac{50}{n} \sum_{l=1}^{n} \left(\frac{l}{n} \right)^{\scriptsize{99}} \longrightarrow \int_{0}^{1} {50x^{\scriptsize{99}} dx} = \frac{1}{2} \\[ 10pt ] &\text{したがって} \\[ 5pt ] &\quad \lim_{n \to \infty} \sum_{k=1}^{2n} (-1)^{\scriptsize{k}} \ \left(\frac{k}{2n} \right)^{\scriptsize{100}} = \frac{1}{2} \end{align*}