2016年入試問題研究に戻る大分大解答
(1) \begin{eqnarray*} R_n-rR_n&=&\sum_{k=0}^{n}(r^k-r^{k+1})=1-r^{n+1}\\ S_n-rS_n&=& \sum_{k=0}^{n}kr^{k-1}-\sum_{k=0}^{n}kr^k= 1+\sum_{k=2}^{n}kr^{k-1}-\sum_{k=1}^{n-1}kr^k-nr^n\\ &=&1+\sum_{k=2}^{n}kr^{k-1}-\sum_{k=2}^{n}(k-1)r^{k-1}-nr^n\\ &=&1+\sum_{k=2}^{n}r^{k-1}-nr^n=R_{n-1}-nr^n \end{eqnarray*} $ r\ne 1 $ なので \begin{eqnarray*} R_n&=&\dfrac{1-r^{n+1}}{1-r},\ \\ S_n&=&\dfrac{1}{1-r}\left(\dfrac{1-r^n}{1-r}-nr^n \right) =\dfrac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2} \end{eqnarray*}
(2) \begin{eqnarray*} T_n-rT_n&=& \sum_{k=2}^{n}k(k-1)r^{k-2}-\sum_{k=2}^{n}k(k-1)r^{k-1}\\ &=&2+\sum_{k=2}^{n-1}k(k+1)r^{k-1}-\sum_{k=2}^{n-1}k(k-1)r^{k-1}-n(n-1)r^{n-1}\\ &=&2+2\sum_{k=2}^{n-1}kr^{k-1}-n(n-1)r^{n-1}=2\sum_{k=1}^{n-1}kr^{k-1}-n(n-1)r^{n-1}\\ &=&2S_{n-1}-n(n-1)r^{n-1} \end{eqnarray*} よって \begin{eqnarray*} T_n&=&\dfrac{1}{1-r}\left(2\cdot\dfrac{1-nr^{n-1}+(n-1)r^n}{(1-r)^2}-n(n-1)r^{n-1} \right)\\ &=&\dfrac{2-(n^2+n)r^{n-1}+2(n^2-1)r^n-(n^2-n)r^n+1}{(1-r)^3} \end{eqnarray*}
(3) ただし書きより \begin{eqnarray*} \lim_{n \to \infty}R_n&=&\dfrac{1}{1-r}\\ \lim_{n \to \infty}S_n&=&\dfrac{1}{(1-r)^2}\\ \lim_{n \to \infty}T_n&=&\dfrac{2}{(1-r)^3} \end{eqnarray*} である．ここで \[ k^2=(k+2)(k+1)-3(k+1)+1 \] なので， \[ \sum_{k=0}^{N}k^2r^k=\sum_{k=0}^{N}\{(k+2)(k+1)r^k-3(k+1)r^k+r^k\} =T_{N+2}-3S_{n+1}+R_N \] よって \[ \sum_{k=0}^{\infty}k^2r^k= \dfrac{2}{(1-r)^3}-\dfrac{3}{(1-r)^2}+\dfrac{1}{1-r} =\dfrac{r^2+r}{(1-r)^3} \]
$ S_n $ , $ T_n $ の求め方2
$ x=\ne 1 $ のとき， \[ 1+x+x^2+\cdots+x^k+\cdots+x^n=\dfrac{1-x^{n+1}}{1-x} \] 両辺を順次微分して, \[ \begin{array}{l} 0+1+2x+\cdots+kx^{k-1}+\cdots+nx^{n-1}= \left(\dfrac{1-x^{n+1}}{1-x} \right)'\\ 0+0+2++\cdots+k(k-1)x^{k-2}+\cdots+n(n-1)x^{n-2} =\left(\dfrac{1-x^{n+1}}{1-x} \right)'' \end{array} \] これを計算して， $ x=r $ を代入そてもよい．