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Question Number 182835 by SANOGO last updated on 15/Dec/22

$$s\left(n\right)=\underset{n=1}{\sum ^{+oo}}\frac{{(-1)}^n+n^2}{n!}{x}^n=?$$

Answered by mr W last updated on 15/Dec/22

$$e^x-1=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$\Rightarrow e^{-x}-1=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^n}{n!}...\left(i\right)$$$$e^x=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{nx}^{n-1}}{n!}$$$${xe}^x=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{nx}^n}{n!}$$$$e^x+{xe}^x=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2{x}^{n-1}}{n!}$$$$\Rightarrow {xe}^x+x^2{e}^x=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2{x}^n}{n!}...\left(ii\right)$$$$\left(i\right)+\left(ii\right):$$$$\Rightarrow s\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n+n^2}{n!}{x}^n=x(1+x)e^x+e^{-x}-1$$

Commented by SANOGO last updated on 15/Dec/22

$$mercibien$$

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