가중평균 수열 등차 증명

$2 + 3 \cdot 2 + \cdots + n ( n - 1 )$ $= 2 ( 2 - 1 ) + 3 ( 3 - 1 ) + \cdots + n ( n - 1 )$ $= 1 ^{2} + 2 ^{2} + 3 ^{2} + \cdots + n ^{2} - ( 1 + 2 + 3 + \cdots + n )$ $\displaystyle = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} - \frac{n ( n + 1 )}{2}$ ∴ (가) $\displaystyle = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ $\displaystyle \frac{2 d \left\{ \frac{n ( n + 1 ) ( 2 n + 1 )}{6} - \frac{n ( n + 1 )}{2} \right\}}{n ( n + 1 )}$ $\displaystyle = {2 d \left\{ \frac{2 n + 1}{6} - \frac{1}{2} \right\}}$ $\displaystyle = \frac{2 d ( 2 n + 1 - 3 )}{6} = \frac{2}{3} d ( n - 1 )$ ∴ (나)$\displaystyle = \frac{2}{3} d$ $\displaystyle \frac{a _{1} + 2 a _{2} + \cdots + na _{n}}{1 + 2 + \cdots + ( n + 1 )}$ $\displaystyle = \frac{1 + 2 + \cdots + n}{1 + 2 + \cdots + ( n + 1 )} \cdot \frac{a _{1} + 2 a _{2} + \cdots + na _{n}}{1 + 2 + \cdots + n}$ $\displaystyle = \frac{\frac{n ( n + 1 )}{2}}{\frac{( n + 1 ) ( n + 2 )}{2}} \cdot b _{n}$ $\displaystyle = \frac{n}{n + 2} \cdot b _{n}$ ∴ (다)$\displaystyle = \frac{n}{n + 2}$ 〔참고〕 주어진 증명을 완성하면 다음과 같다. $\displaystyle b _{n+1} = \frac{n}{n + 2} b _{n} + \frac{2}{n + 2} a _{n+1}$ $\displaystyle = \left( 1 - \frac{2}{n + 2} \right) b _{n} + \frac{2}{n + 2} a _{n+1}$ $\displaystyle = b _{n} - \frac{2}{n + 2} b _{n} + \frac{2}{n + 2} a _{n+1}$ ∴ $\displaystyle b _{n+1} - b _{n} = \frac{2}{n + 2} ( a _{n+1} - b _{n} )$ ⋯ ㉠ 그런데, 등차수열 $\{ b _{n} \}$의 첫째항을 $a'$, 공차를 $d'$이라 하면 $b _{n+1} - b _{n} = d'$이므로 ㉠에서 $\displaystyle d' = \frac{2}{n + 2} ( a _{n+1} - b _{n} )$ ∴ $\displaystyle a _{n+1} = \frac{d'}{2} ( n + 2 ) + b _{n}$ $\displaystyle = \frac{d'}{2} ( n + 2 ) + a' + ( n - 1 ) d'$ $\displaystyle = \left( a' + \frac{3}{2} d' \right) + \frac{3}{2} d' ( n - 1 )$ 따라서 수열 {$a _{n+1}$}은 $\displaystyle a _{2} = a' + \frac{3}{2} d'$이고 공차가 $\displaystyle \frac{3}{2} d'$인 등차수열이다. 그런데, $a _{1} = b _{1}$$= a'$이므로 수열 $\{ a _{n} \}$은 첫째항이 $a'$, 공차가 $\displaystyle \frac{3}{2} d'$인 등차수열이다.