求无穷级数之和 ∑(n=1,∞)n(n+1)／2^n ，∑(n=1,∞)n(n+1)(n+2)／2^n ，……

wintex

請問級數

luyuanhong

fungarwai

個方法係將組合數換成多項式入面嘅係數，求等比數列和，再攞番和式入面果個係數

\(\displaystyle \frac{n(n+1)}{2^n}=2!\binom{n+1}{2}\frac{1}{2^n}
=2![x^2]\frac{(1+x)^{n+1}}{2^n}
=4[x^2]\left(\frac{1+x}{2}\right)^{n+1}\)

\(\displaystyle \sum_{n=1}^\infty \frac{n(n+1)}{2^n}
=4[x^2]\sum_{n=1}^\infty \left(\frac{1+x}{2}\right)^{n+1}
=[x^2]\frac{(1+x)^2}{1-(1+x)/2}
=2[x^2]\frac{(1+x)^2}{1-x}\)

\(\displaystyle =2[x^2](1+2x+x^2)\sum_{n=0}^\infty x^n
=2(1+2+1)=8\)


\(\displaystyle \frac{n(n+1)(n+2)\dots(n+k-1)}{2^n}=k!\binom{n+k-1}{k}\frac{1}{2^n}
=k!2^{k-1}[x^k]\left(\frac{1+x}{2}\right)^{n+k-1}\)

\(\displaystyle k!2^{k-1}[x^k]
\sum_{n=1}^\infty \left(\frac{1+x}{2}\right)^{n+k-1}
=k![x^k]\frac{(1+x)^k}{1-x}
=k^k\sum_{n=0}^\infty x^n
=k!2^k\)