分母に階乗の和を含む総和
分母に階乗の和を含む総和
\[ \frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\cdots+\frac{100}{98!+99!+100!} \] を求めよ。
\[ \frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\cdots+\frac{100}{98!+99!+100!} \] を求めよ。
\begin{align*}
\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\cdots+\frac{100}{98!+99!+100!} & =\sum_{k=1}^{98}\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}\\
& =\sum_{k=1}^{98}\frac{k+2}{k!\left(1+k+1+\left(k+1\right)\left(k+2\right)\right)}\\
& =\sum_{k=1}^{98}\frac{k+2}{k!\left(k^{2}+4k+4\right)}\\
& =\sum_{k=1}^{98}\frac{1}{k!\left(k+2\right)}\\
& =\sum_{k=1}^{98}\frac{k+1}{\left(k+2\right)!}\\
& =\sum_{k=1}^{98}\frac{k+2-1}{\left(k+2\right)!}\\
& =\sum_{k=1}^{98}\frac{1}{\left(k+1\right)!}-\frac{1}{\left(k+2\right)!}\\
& =\frac{1}{2}-\frac{1}{100!}
\end{align*}
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