積分漸化式

$n=0,1,2,\cdots$に対し$\{a_n\}$と$\{b_n\}$を次のように定義する。

${\displaystyle a_n=\frac{1}{\pi }{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1+\sin x{)}^{n}dx}$
$b_n=a_{n+1}-a_n$

この時次の問いに答えよ

(1)$a_n$を$n$を用いて表せ。
(2)$b_n$を$n$を用いて表せ。
(3)${\displaystyle S_n=\sum _{k=0}^n\frac{1}{2^k}{}_{2k}{\mathrm{C}}_{k+1}}$を$n$を用いて表せ。

解答
(1)
${\displaystyle I_n={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1+\sin x{)}^{n}dx}$とすると

${\displaystyle I_n={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1+\sin x)(1+\sin x{)}^{n-1}dx}$
${\displaystyle ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1+\sin x{)}^{n-1}dx+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin x(1+\sin x{)}^{n-1}dx}$

ここで${\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin x(1+\sin x{)}^{n-1}dx=J_{n-1}}$とすれば

$I_n=I_{n-1}+J_{n-1}\cdots$①

また$J_{n-1}$を部分積分することで

${\displaystyle J_{n-1}}$
${\displaystyle =[-\cos x(1+\sin x{)}^{n-1}{]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-\cos x\{(1+\sin x{)}^{n-1}{\}}^{\prime }dx}$
${\displaystyle =(n-1){\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^{2}x(1+\sin x{)}^{n-2}dx}$
${\displaystyle =(n-1){\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1-{\sin }^{2}x)(1+\sin x{)}^{n-2}dx}$
${\displaystyle =(n-1){\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1-\sin x)(1+\sin x)(1+\sin x{)}^{n-2}dx}$
${\displaystyle =(n-1){\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1-\sin x)(1+\sin {)}^{n-1}dx}$
${\displaystyle =(n-1)(I_{n-1}-J_{n-1})\cdots }$②

②より${\displaystyle J_{n-1}=\frac{n-1}{n}{I}_{n-1}\cdots }$③

①に③を代入することで

${\displaystyle I_n=\frac{2n-1}{n}{I}_{n-1}\cdots }$④

④を繰り返し用いることで

${\displaystyle I_n=\frac{2n-1}{n}\cdot \frac{2n-3}{n-1}\cdots \frac{3}{2}\cdot \frac{1}{1}\cdot I_0}$

ここで${\displaystyle I_0={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}dx=\pi }$より

${\displaystyle I_n=\frac{(2n-1)!!}{n!}\pi }$
${\displaystyle =\frac{(2n-1)!}{2^{n-1}(n-1)!n!}\pi }$
${\displaystyle =\frac{1}{2^{n-1}}{}_{2n-1}{\mathrm{C}}_{n-1}\pi }$

${\displaystyle a_n=\frac{1}{\pi }{I}_n}$より

${\displaystyle a_n=\frac{1}{2^{n-1}}{}_{2n-1}{\mathrm{C}}_{n-1}}$

(2)
$b_n=a_{n+1}-a_n$より

${\displaystyle b_n=\frac{1}{2^n}{}_{2n}{\mathrm{C}}_{n+1}}$

(3)
(2)より
${\displaystyle S_n=\sum _{k=0}^n{b}_k}$
${\displaystyle =(a_{n+1}-a_n)+(a_n-a_{n-1})+\cdots +(a_1-a_0)}$
${\displaystyle =a_{n+1}-a_0}$
${\displaystyle =\frac{1}{2^n}{}_{2n+1}{\mathrm{C}}_n-1}$