上級者
数学Ⅱ：式の証明
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ヒント：(2)の問題は(1)の結果を使えることが多い．
問題
\(n\)を自然数とするとき，次の和を求めよ．
(A) \(\displaystyle \sum_{k=0}^{n} k {}_{n}C_{k}\)
(B) \(\displaystyle \sum_{k=0}^{n} k^{2} {}_{n}C_{k}\)
(C) \(\displaystyle \sum_{k=0}^{n} k^{3} {}_{n}C_{k}\)
解答
(A)\(k{}_{n}C_{k}=n{}_{n-1}C_{k-1}\)(\(n \geqq 1,~k \geqq 1\))なので，
\(\displaystyle \sum_{k=0}^{n} k {}_{n}C_{k}\)\(=\displaystyle \sum_{k=1}^{n} k {}_{n}C_{k}\)\(=\displaystyle n\sum_{k=1}^{n} {}_{n-1}C_{k-1}\)\(=\displaystyle n\sum_{k=0}^{n-1} {}_{n-1}C_{k}\)
ここで二項定理より，
\((1+x)^{n}\)\(=\displaystyle \sum_{k=0}^{n} {}_{n}C_{k}x^{k}\)
\(x=1\), \(n\rightarrow n-1\)とすると，
\(\displaystyle \sum_{k=0}^{n-1} {}_{n-1}C_{k}\)\(=2^{n-1}\)
よって，\(\displaystyle \sum_{k=0}^{n} k {}_{n}C_{k}\)\(=n2^{n-1}\)
(B)
\(\displaystyle \sum_{k=0}^{n} k^{2} {}_{n}C_{k}\)\(=\displaystyle \sum_{k=1}^{n} k\cdot k{}_{n}C_{k}\)\(=\displaystyle \sum_{k=1}^{n} kn{}_{n-1}C_{k-1}\)\(=\displaystyle n\sum_{k=1}^{n} (k-1){}_{n-1}C_{k-1}\)\(+\displaystyle n\sum_{k=1}^{n} {}_{n-1}C_{k-1}\)\(=\displaystyle n\sum_{k=0}^{n-1} k {}_{n-1}C_{k}\)\(+\displaystyle n\sum_{k=0}^{n-1} {}_{n-1}C_{k}\)
ここで(A)より，\(\displaystyle \sum_{k=0}^{n-1} k {}_{n-1}C_{k}\)\(=(n-1)2^{n-2}\) \((n\geqq 2)\)
また，\(\displaystyle n\sum_{k=0}^{n-1} {}_{n-1}C_{k}\)\(=2^{n-1}\)
よって，\(\displaystyle \sum_{k=0}^{n} k^{2} {}_{n}C_{k}\)\(=n(n-1)2^{n-2}\)\(+n2^{n-1}\)\(=n(n+1)2^{n-2}\)
これは，\(n=1\)でも成立する．
(C)
\(\displaystyle \sum_{k=0}^{n} k^{3} {}_{n}C_{k}\)\(=\displaystyle \sum_{k=1}^{n} k^{2}\cdot k{}_{n}C_{k}\)\(=\displaystyle \sum_{k=1}^{n} k^{2} n{}_{n-1}C_{k-1}\)\(=\displaystyle n\sum_{k=1}^{n} (k-1)^{2} {}_{n-1}C_{k-1}\)\(+2\displaystyle n\sum_{k=1}^{n} (k-1) {}_{n-1}C_{k-1}\)\(+\displaystyle n\sum_{k=1}^{n} {}_{n-1}C_{k-1}\)\(=\displaystyle n\sum_{k=0}^{n-1} k^{2} {}_{n-1}C_{k}\)\(+2\displaystyle n\sum_{k=0}^{n-1} k {}_{n-1}C_{k}\)\(+\displaystyle n\sum_{k=0}^{n-1} {}_{n-1}C_{k}\)
ここで，(B)より，\(\displaystyle \sum_{k=0}^{n-1} k^{2} {}_{n-1}C_{k}\)\(=(n-1)n2^{n-3}\) \((n\geqq 2)\)
また，\(\displaystyle \sum_{k=0}^{n-1} k {}_{n-1}C_{k}\)\(=(n-1)2^{n-2}\) \((n\geqq 2)\)
また，\(\displaystyle \sum_{k=0}^{n-1} {}_{n-1}C_{k}\)\(=2^{n-1}\)
よって，\(\displaystyle \sum_{k=0}^{n} k^{3} {}_{n}C_{k}\)\(=n^2(n+3)2^{n-3}\)
これは，\(n=1,2\)でも成立する．
参考入試問題
（１）\(n\)を自然数とする．次の等式が成り立つように定数\(a,b\)を定めよ．
\(\displaystyle \frac{n+1}{y(y+1)\cdots(y+n)(y+n+1)}\)\(=\displaystyle \frac{a}{y(y+1)\cdots(y+n)}\)+\(\displaystyle \frac{b}{(y+1)\cdots(y+n+1)}\)
（２）すべての自然数\(n\)について，次の等式が成り立つことを証明せよ．
\(\displaystyle \frac{n!}{x(x+1)\cdots(x+n)}\)\(=\displaystyle \sum_{r=0}^{n} (-1)^{r}\frac{{}_{n}C_{r}}{x+r}\) （2006前期大阪大）
解答
(1)
\(\displaystyle \frac{a}{y(y+1)\cdots(y+n)}\)+\(\displaystyle \frac{b}{(y+1)\cdots(y+n+1)}\)\(=\displaystyle \frac{a(y+n+1)+by}{y(y+1)\cdots(y+n)(y+n+1)}\)\(=\displaystyle \frac{(a+b)y+a(n+1)}{y(y+1)\cdots(y+n)(y+n+1)}\)
よって\(a+b=0,a=1\)
∴\(a=1,b=-1\)
(2)
数学的帰納法で証明する．
\(n=1\)のとき
右辺\(=\displaystyle \sum_{r=0}^{n} (-1)^{r}\frac{{}_{n}C_{r}}{(x+r)}\)\(=\displaystyle \frac{{}_{1}C_{0}}{x}-\frac{{}_{1}C_{1}}{x+1}\)\(=\displaystyle \frac{1}{x(x+1)}\)
左辺\(=\displaystyle \frac{1}{x(x+1)}\)
よって成立する．
\(n=k\)のとき成り立つとすると，
まず(1)から
\(\displaystyle \frac{k+1}{x(x+1)\cdots(x+k)(x+k+1)}\)\(=\displaystyle \frac{1}{x(x+1)\cdots(x+k)}\)-\(\displaystyle \frac{1}{(x+1)\cdots(x+k+1)}\)
両辺に\(k!\)をかけると，
\(\displaystyle \frac{(k+1)!}{x(x+1)\cdots(x+k)(x+k+1)}\)\(=\displaystyle \frac{k!}{x(x+1)\cdots(x+k)}\)\(-\displaystyle \frac{k!}{(x+1)\cdots(x+k+1)}\)\(=\displaystyle \sum_{r=0}^{k} (-1)^{r}\frac{{}_{k}C_{r}}{x+r}\)\(-\displaystyle \sum_{r=0}^{k} (-1)^{r}\frac{{}_{k}C_{r}}{x+1+r}\)
ここで\(\displaystyle \sum_{r=0}^{n} (-1)^{r}\frac{{}_{k}C_{r}}{x+1+r}\)を\(r \rightarrow r-1\)とすると，\(\displaystyle \sum_{r=1}^{k+1} (-1)^{r-1}\frac{{}_{k}C_{r-1}}{x+r}\)
よって，
\(\displaystyle \sum_{r=0}^{k} (-1)^{r}\frac{{}_{k}C_{r}}{x+r}\)\(-\displaystyle \sum_{r=0}^{k} (-1)^{r}\frac{{}_{k}C_{r}}{x+1+r}\)\(=\displaystyle \frac{{}_{k}C_{0}}{x}\)\(+\displaystyle \sum_{r=1}^{k} (-1)^{r}\frac{{}_{k}C_{r}}{x+r}\)\(+\displaystyle \sum_{r=1}^{k} (-1)^{r}\frac{{}_{k}C_{r-1}}{x+r}\)\(+\displaystyle (-1)^{k+1}\frac{{}_{k+1}C_{k+1}}{x+k+1}\)\(=\displaystyle \frac{1}{x}\)\(+\displaystyle \frac{(-1)^{k+1}}{x+k+1}\)\(+\displaystyle \sum_{r=1}^{k} (-1)^{r}\frac{{}_{k}C_{r}+{}_{k}C_{r-1}}{x+r}\)
ここで\({}_{k}C_{r}+{}_{k}C_{r-1}={}_{k+1}C_{r}\)なので，
与式\(=\displaystyle \frac{1}{x}\)\(+\displaystyle \frac{(-1)^{k+1}}{x+k+1}\)\(+\displaystyle \sum_{r=1}^{k} (-1)^{r}\frac{{}_{k+1}C_{r}}{x+r}\)\(=\displaystyle \frac{(-1)^{0}{}_{k}C_{0}}{x+0}\)\(+\displaystyle \sum_{r=1}^{k} (-1)^{r}\frac{{}_{k+1}C_{r}}{x+r}\)\(+\displaystyle \frac{(-1)^{k+1}{}_{k+1}C_{k+1}}{x+k+1}\)\(=\displaystyle \sum_{r=0}^{k+1} (-1)^{r}\frac{{}_{k+1}C_{r}}{x+r}\)
よって\(n=k+1\)でも成り立つ．
以上数学的帰納法より\(\displaystyle \frac{n!}{x(x+1)\cdots(x+n)}\)\(=\displaystyle \sum_{r=0}^{n} (-1)^{r}\frac{{}_{n}C_{r}}{x+r}\)が成り立つ．