化簡一下...拆項對消即可
\( \displaystyle \frac{k}{C_3^{k+2}}=\frac{k}{\frac{(k+2)!}{(k-1)!3!}}=\frac{6}{(k+1)(k+2)}=6 \left( \frac{1}{k+1}-\frac{1}{k+2} \right) \)
原式\( \displaystyle =6 \sum_{k=1}^{\infty} \left( \frac{1}{k+1}-\frac{1}{k+2} \right)=\ldots=3 \)
