$\sum_{k=0}^{n} \frac{(-1)^k {}_n C_k}{(k+1)(k+3)}$ を $n$ の式で表す。

1. 問題の内容

$\sum_{k=0}^{n} \frac{(-1)^k {}_n C_k}{(k+1)(k+3)}$ を $n$ の式で表す。

2. 解き方の手順

まず、$\frac{1}{(k+1)(k+3)}$ を部分分数分解します。

$$\frac{1}{(k+1)(k+3)} = \frac{A}{k+1} + \frac{B}{k+3}$$

$1 = A(k+3) + B(k+1)$ を満たす $A$ と $B$ を求めます。

$k = -1$ のとき $1 = 2A$ なので $A = \frac{1}{2}$.

$k = -3$ のとき $1 = -2B$ なので $B = -\frac{1}{2}$.

したがって、

$$\frac{1}{(k+1)(k+3)} = \frac{1}{2} \left( \frac{1}{k+1} - \frac{1}{k+3} \right)$$

与えられた式は

$$\sum_{k=0}^{n} \frac{(-1)^k {}_n C_k}{(k+1)(k+3)} = \frac{1}{2} \sum_{k=0}^{n} (-1)^k {}_n C_k \left( \frac{1}{k+1} - \frac{1}{k+3} \right)$$

ここで、${}_n C_k = \frac{n!}{k!(n-k)!}$ を用いて

$$\frac{{}_n C_k}{k+1} = \frac{n!}{(k+1)!(n-k)!} = \frac{1}{n+1} \frac{(n+1)!}{(k+1)!((n+1)-(k+1))!} = \frac{1}{n+1} {}_{n+1} C_{k+1}$$

同様に

$$\frac{{}_n C_k}{k+3} = \frac{n!}{(k+3)!(n-k)!} = \frac{n!}{(k+3)!(n-k)!} \frac{(n+1)(n+2)(n+3)}{(n+1)(n+2)(n+3)} = \frac{(n+3)!}{(k+3)!(n-k)!(n+1)(n+2)(n+3)} = \frac{{}_{n+3}C_{k+3}}{(n+1)(n+2)(n+3)}$$

$$\frac{{}_n C_k}{k+3} = \frac{6}{(n+1)(n+2)(n+3)} \frac{(n+3)!}{6(k+3)!(n-k)!}$$

ここで、

$$\frac{{}_n C_k}{k+3} = \frac{{}_{n+3} C_{k+3} \frac{(k+3)!(n-k)!}{(n+3)!} n!}{(n+1)(n+2)(n+3)}$$

次に

$$\sum_{k=0}^{n} \frac{(-1)^k {}_{n+1} C_{k+1}}{n+1} = \frac{1}{n+1} \sum_{k=0}^{n} (-1)^k {}_{n+1} C_{k+1}$$

$j = k+1$ とすると $k = j-1$ であり $k=0$ のとき $j=1$ , $k=n$ のとき $j=n+1$.

$$\frac{1}{n+1} \sum_{j=1}^{n+1} (-1)^{j-1} {}_{n+1} C_{j} = \frac{-1}{n+1} \sum_{j=1}^{n+1} (-1)^{j} {}_{n+1} C_{j} = \frac{-1}{n+1} \left( \sum_{j=0}^{n+1} (-1)^{j} {}_{n+1} C_{j} - {}_{n+1} C_0 \right)$$

二項定理より、$\sum_{j=0}^{n+1} (-1)^{j} {}_{n+1} C_{j} = (1-1)^{n+1} = 0$ なので、

$$\frac{-1}{n+1}(0 - 1) = \frac{1}{n+1}$$

$$\sum_{k=0}^{n} \frac{(-1)^k {}_{n+3} C_{k+3}}{(n+1)(n+2)(n+3)} = \frac{1}{(n+1)(n+2)(n+3)} \sum_{k=0}^{n} (-1)^k {}_{n+3} C_{k+3}$$

$j = k+3$ とおくと $k = j-3$. $k=0$ のとき $j=3$, $k=n$ のとき $j=n+3$

$$= \frac{1}{(n+1)(n+2)(n+3)} \sum_{j=3}^{n+3} (-1)^{j-3} {}_{n+3} C_{j} = \frac{1}{(n+1)(n+2)(n+3)} \sum_{j=3}^{n+3} (-1)^{j-3} {}_{n+3} C_{j}$$

$$(-1)^k = (-1)^{j-3} = (-1)^j (-1)^{-3} = - (-1)^j$$

$$\sum_{j=3}^{n+3} (-1)^{j-3} {}_{n+3} C_{j} = - \sum_{j=3}^{n+3} (-1)^{j} {}_{n+3} C_{j} = - \left( \sum_{j=0}^{n+3} (-1)^{j} {}_{n+3} C_{j} - {}_{n+3} C_{0} + {}_{n+3} C_{1} - {}_{n+3} C_{2} \right)$$

$$= - \left( 0 - 1 + (n+3) - \frac{(n+3)(n+2)}{2} \right) = 1 - (n+3) + \frac{(n+3)(n+2)}{2} = 1 - n - 3 + \frac{n^2+5n+6}{2} = -n - 2 + \frac{n^2+5n+6}{2} = \frac{-2n - 4 + n^2 + 5n + 6}{2} = \frac{n^2 + 3n + 2}{2} = \frac{(n+1)(n+2)}{2}$$

$$\frac{1}{2} \left(\frac{1}{n+1} - \frac{(n+1)(n+2)}{2(n+1)(n+2)(n+3)}\right) = \frac{1}{2} \left( \frac{1}{n+1} - \frac{1}{2(n+3)}\right) = \frac{1}{2} \left(\frac{2n+6-n-1}{2(n+1)(n+3)} \right) = \frac{1}{2} \frac{n+5}{2(n+1)(n+3)} = \frac{n+5}{4(n+1)(n+3)}$$

$$\sum_{k=0}^{n} \frac{(-1)^k {}_n C_k}{(k+1)(k+3)} = \frac{n+5}{4(n+1)(n+3)}$$

$$\frac{1}{2} \sum_{k=0}^n (-1)^k \binom{n}{k} \left( \frac{1}{k+1} - \frac{1}{k+3} \right) = \frac{1}{2} \left(\frac{1}{n+1} + \frac{(n+1)(n+2)}{2(n+1)(n+2)(n+3)}\right)$$

$$= \frac{1}{2(n+1)} - \frac{1}{2} \frac{\sum_k (-1)^k \binom{n}{k}}{k+3}$$

部分分数分解により、与式は

$$\frac{1}{2} \sum_{k=0}^n (-1)^k \binom{n}{k} (\frac{1}{k+1}-\frac{1}{k+3})$$

$$\sum_{k=0}^n \frac{(-1)^k \binom{n}{k}}{k+1} = \frac{1}{n+1}$$

$$\sum_{k=0}^n \frac{(-1)^k \binom{n}{k}}{k+3} = \frac{2}{(n+1)(n+2)(n+3)}$$

よって

$$\frac{1}{2} [\frac{1}{n+1} - \frac{2}{(n+1)(n+2)(n+3)}] = \frac{1}{2} [\frac{(n+2)(n+3)-2}{(n+1)(n+2)(n+3)}] = \frac{n^2+5n+4}{2(n+1)(n+2)(n+3)} = \frac{(n+4)(n+1)}{2(n+1)(n+2)(n+3)} = \frac{n+4}{2(n+2)(n+3)}$$

3. 最終的な答え

$\frac{n+4}{2(n+2)(n+3)}$