$$ { 求级数:\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n\left( n^2-n+1 \right)}{2^n}}的和.} $$
Tips Answer$$ { 解:\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n\left( n^2-2n+1 \right)}{2^n}}\begin{array}{l} =\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n\left[ n\left( n-1 \right) +1 \right]}{2^n}}\\ =\sum_{n=2}^{\infty}{n\left( n-1 \right) \frac{\left( -1 \right) ^n}{2^n}}+\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{2^n}}\\ =\sum_{n=2}^{\infty}{n\left( n-1 \right) \left( -\frac{1}{2} \right) ^n+\sum_{n=0}^{\infty}{\left( -\frac{1}{2} \right) ^n}};\\ \end{array}} \\ { \sum_{n=0}^{\infty}{\left( -\frac{1}{2} \right) ^n}=\frac{1}{1-\left( -\frac{1}{2} \right)}=\frac{1}{1+\frac{1}{2}}=\frac{1}{\frac{3}{2}}=\frac{2}{3}.} \\ { 由于\sum_{n=2}^{\infty}{n\left( n-1 \right) \left( -\frac{1}{2} \right) ^n},因此我们设S\left( x \right) =\sum_{n=2}^{\infty}{n\left( n-1 \right) x^{n-2}};于是有:} \\ { S\left( x \right) \begin{array}{l} =\sum_{n=2}^{\infty}{n\left( n-1 \right) x^{n-2}}=S\left( x \right) =\left( \sum_{n=0}^{\infty}{x^n} \right) ^{''}=\left( \frac{1}{1-x} \right) ^{''}=\left[ \frac{-\left( -1 \right)}{\left( 1-x \right) ^2} \right] ^{\prime}\\ =\frac{-2\left( 1-x \right) \cdot \left( -1 \right)}{\left( 1-x \right) ^4}=\frac{2}{\left( 1-x \right) ^3};其中收敛域为\left| x \right|<1.\\ \end{array}} \\ { 于是x^2S\left( x \right) =x^2\cdot \sum_{n=2}^{\infty}{n\left( n-1 \right) x^{n-2}}=\frac{2x^2}{\left( 1-x \right) ^3},\left| x \right|<1.} \\ { 于是x^2S\left( \frac{1}{2} \right) =\sum_{n=2}^{\infty}{n\left( n-1 \right) \left( -\frac{1}{2} \right) ^n}=\frac{2\cdot \left( -\frac{1}{2} \right) ^2}{\left[ 1-\left( -\frac{1}{2} \right) \right] ^3}=\frac{2\cdot \frac{1}{4}}{\left( \frac{3}{2} \right) ^3}=\frac{\frac{1}{2}}{\frac{27}{8}}=\frac{1}{2}\cdot \frac{8}{27}=\frac{4}{27}.} \\ { \sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n\left( n^2-n+1 \right)}{2^n}}=\frac{4}{27}+\frac{2}{3}=\frac{4}{27}+\frac{18}{27}=\frac{22}{27}.} \\ { 注1:为什么不是S\left( x \right) =\sum_{n=0}^{\infty}{n\left( n-1 \right) \left( -\frac{1}{2} \right) ^n}} \\ { 因为第0项和第1项均为0,所以从第2项开始.} \\ { 注2:为什么设S\left( x \right) =\sum_{n=2}^{\infty}{n\left( n-1 \right) x^{n-2}}} \\ { 因为我们熟知\sum_{n=0}^{\infty}{x^n}=\frac{1}{1-x},\left| x \right|<1,我们要想办法找到这个熟悉的结构;} \\ { 我们发现n\left( n-1 \right) 这个结构是对\sum_{n=0}^{\infty}{x^n}求两次求导得到的,因此我们设为x^{n-2};} \\ { 这样最后只需要补一个x^2就得到\sum_{n=2}^{\infty}{n\left( n-1 \right) x^n}这个结构了.} \\ $$