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Question Number 73231 by mathmax by abdo last updated on 08/Nov/19

$$findthesumof\sum _{n=0}^{\mathrm{\infty }}(n^2-3n+1)e^{-n}$$

Commented by mathmax by abdo last updated on 09/Nov/19

$$wehave\sum _{n=0}^{\mathrm{\infty }}(n^2-3n+1)e^{-n}=\sum _{n=0}^{\mathrm{\infty }}{n}^2{\left(e^{-1}\right)}^n-3\sum _{n=0}^{\mathrm{\infty }}n{\left(e^{-1}\right)}^n$$$$+\sum _{n=0}^{\mathrm{\infty }}{e}^{-n}first\sum _{n=0}^{\mathrm{\infty }}{e}^{-n}=\sum _{n=0}^{\mathrm{\infty }}{\left(e^{-1}\right)}^n=\frac{1}{1-e^{-1}}=\frac{e}{e-1}$$$$letcalculatef\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{nx}^{n}andg\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{n}^2{x}^{n}with\mid x\mid <1$$$$wehave\sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{nx}^{n-1}=\frac{1}{{(1-x)}^2}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}n{x}^n=\frac{x}{{(1-x)}^2}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}{ne}^{-n}=\frac{e^{-1}}{{(1-e^{-1})}^2}=\frac{e^2}{{(e-1)}^2}also$$$$\sum _{n=1}^{\mathrm{\infty }}{n}^2{x}^{n-1}=\frac{{(1-x)}^2-x(-2)(1-x)}{{(1-x)}^4}=\frac{1-x+2x}{{(1-x)}^3}=\frac{x+1}{{(1-x)}^3}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}{n}^2{x}^n=\frac{x^2+x}{{(1-x)}^3}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}{n}^2{e}^{-n}=\frac{e^{-2}+e^{-1}}{{(1-e^{-1})}^3}$$$$=\frac{e+e^2}{{(e-1)}^3}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}(n^2-3n+1)e^{-n}=\frac{e+e^2}{{(e-1)}^3}-\frac{3e^2}{{(e-1)}^2}+\frac{e}{e-1}$$$$=\frac{e+e^2-3e^2(e-1)+e{(e-1)}^2}{{(e-1)}^3}=\frac{e+e^2-3e^3+3e^2+e(e^2-2e+1)}{{(e-1)}^3}$$$$=\frac{-3e^3+4e^2+e+e^3-2e^2+e}{{(e-1)}^3}=\frac{-2e^3+2e^2+2e}{{(e-1)}^3}$$

Answered by mind is power last updated on 08/Nov/19

$${\mathrm{n}}^2-3\mathrm{n}+1=(\mathrm{n}+1)(\mathrm{n}+2)-6(\mathrm{n}+1)+5$$$$\sum _{\mathrm{n}⩾0}(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{e}}^{-\mathrm{n}}-6\sum _{\mathrm{n}⩾0}(\mathrm{n}+1){\mathrm{e}}^{-\mathrm{n}}+5\sum _{\mathrm{n}⩾0}{\mathrm{e}}^{-\mathrm{n}}$$$$\frac{1}{1-\mathrm{x}}=\mathrm{\Sigma }{\mathrm{x}}^{\mathrm{k}}\Rightarrow \frac{1}{{(1-\mathrm{x})}^2}.\sum _{\mathrm{k}⩾1}{\mathrm{kx}}^{\mathrm{k}-1}=\sum _{\mathrm{k}⩾0}(\mathrm{k}+1){\mathrm{x}}^{\mathrm{k}}$$$$\mathrm{x}={\mathrm{e}}^{-1}\Rightarrow \mathrm{\Sigma }{\mathrm{e}}^{-\mathrm{n}}=\frac{1}{1-{\mathrm{e}}^{-1}}=\frac{\mathrm{e}}{\mathrm{e}-1}$$$$\Rightarrow \mathrm{x}={\mathrm{e}}^{-1}\Rightarrow \frac{1}{{(1-{\mathrm{e}}^{-1})}^2}=\sum _{\mathrm{n}⩾0}(\mathrm{n}+1){\mathrm{e}}^{-\mathrm{n}}=\frac{{\mathrm{e}}^2}{(\mathrm{e}-1)}$$$$\frac{1}{{(1-\mathrm{x})}^2}=\sum _{\mathrm{k}⩾0}(\mathrm{k}+1){\mathrm{x}}^{\mathrm{k}}\Rightarrow \sum _{\mathrm{k}⩾1}\mathrm{k}(\mathrm{k}+1){\mathrm{x}}^{\mathrm{k}-1}=\frac{2}{{(1-\mathrm{x})}^3}=\sum _{\mathrm{k}⩾0}(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{x}}^{\mathrm{k}}$$$$\mathrm{x}={\mathrm{e}}^{-1}\Rightarrow \sum _{\mathrm{n}⩾0}(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{e}}^{-\mathrm{n}}=\frac{2}{{(1-{\mathrm{e}}^{-1})}^3}=\frac{{2\mathrm{e}}^3}{{(\mathrm{e}-1)}^3}$$$$\mathrm{so}\mathrm{we}\mathrm{get}\frac{{2\mathrm{e}}^3}{{(\mathrm{e}-1)}^3}-\frac{{6\mathrm{e}}^2}{{(\mathrm{e}-1)}^2}+\frac{5\mathrm{e}}{(\mathrm{e}-1)}$$$$=\frac{{\mathrm{e}}^3-{4\mathrm{e}}^2+5\mathrm{e}}{{(\mathrm{e}-1)}^3}$$$$$$

Commented by mathmax by abdo last updated on 08/Nov/19

$$thankxsir.$$

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