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Question Number 71216 by TawaTawa last updated on 13/Oct/19

Answered by mind is power last updated on 13/Oct/19

$$\mathrm{\Sigma }\frac{{\mathrm{n}}^2+\mathrm{x}}{\mathrm{n}!}=\mathrm{\Sigma }\frac{{\mathrm{n}}^2}{\mathrm{n}!}+\mathrm{x}\mathrm{\Sigma }\frac{1}{\mathrm{n}!}$$$$\sum _{\mathrm{n}⩾0}\frac{{\mathrm{n}}^2}{\mathrm{n}!}=1+\sum _{\mathrm{n}⩾2}\frac{\mathrm{n}}{(\mathrm{n}-1)!}=\mathrm{\Sigma }\frac{\mathrm{n}-1+1}{(\mathrm{n}-1)!}=\mathrm{\Sigma }\frac{1}{(\mathrm{n}-2)!}+\mathrm{\Sigma }\frac{1}{(\mathrm{n}-1)!}$$$$=\mathrm{e}+\mathrm{e}-1$$$$\Rightarrow \mathrm{\Sigma }\frac{{\mathrm{n}}^2}{\mathrm{n}!}=1+2\mathrm{e}-1=2\mathrm{e}$$$$\mathrm{\Sigma }\frac{\mathrm{x}}{\mathrm{n}!}=\mathrm{xe}$$$$\Rightarrow 2\mathrm{e}+\mathrm{xe}={\mathrm{ex}}^2$$$$\Rightarrow {\mathrm{x}}^2-\mathrm{x}-2=0$$$$(\mathrm{x}-2)(\mathrm{x}+1)=0$$$$\mathrm{x}\in \{2,-1\}$$$$$$

Commented by TawaTawa last updated on 13/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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