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Question Number 73261 by mathmax by abdo last updated on 09/Nov/19

$$calculate\sum _{n=1}^{\mathrm{\infty }}(n^4+2n^2-3)\frac{x^n}{n!}incaseofconvergence.$$

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

$$wehave\sum _{n=1}^{\mathrm{\infty }}(n^4+2n^2-3)\frac{x^n}{n!}=\sum _{n=1}^{\mathrm{\infty }}\frac{n^3}{(n-1)!}{x}^n+2\sum _{n=1}^{\mathrm{\infty }}\frac{{nx}^n}{(n-1)!}$$$$-3\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n!}wehavefirst\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n!}=e^x-1$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{nx}^n}{(n-1)!}=\sum _{n=0}^{\mathrm{\infty }}\frac{(n+1)x^{n+1}}{n!}=\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n+1}}{(n-1)!}+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{n!}$$$$=x\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{n!}+x{e}^x=x^2{e}^x+{xe}^x$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^3}{(n-1)!}{x}^n=\sum _{n=0}^{\mathrm{\infty }}\frac{{(n+1)}^3}{n!}{x}^{n+1}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{n^3+3n^2+n+1}{n!}{x}^{n+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{(n-1)!}{x}^{n+1}+3\sum _{n=1}^{\mathrm{\infty }}\frac{n}{(n-1)!}{x}^{n+1}$$$$+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n-1)!}+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{n!}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{(n-1)!}{x}^{n+1}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(n+1)}^2{x}^{n+2}}{n!}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{n^2+2n+1}{n!}{x}^{n+2}=\sum _{n=1}^{\mathrm{\infty }}\frac{n}{(n-1)!}{x}^{n+2}+2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n-1)!}{x}^{n+2}$$$$+x^2{e}^x$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{n+1}{n!}{x}^{n+2}+2\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+3}}{n!}+x^2{e}^x$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n+2}}{(n-1)!}+x^2{e}^x+2x^3{e}^x+x^2{e}^x$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+3}}{n!}+2x^2{e}^x+2x^3{e}^x$$$$=x^3{e}^x+2x^3{e}^x+2x^2{e}^x=(3x^3+2x^2)e^x...becontinued...$$

Answered by mind is power last updated on 09/Nov/19

$$\mathrm{use}1,\mathrm{x},\mathrm{x}(\mathrm{x}-1),\mathrm{x}(\mathrm{x}-1)(\mathrm{x}-2),\mathrm{x}(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)$$$$\mathrm{as}\mathrm{Base}\mathrm{of}{\mathrm{IR}}^4\left[\mathrm{X}\right]$$$${\mathrm{n}}^4+{2\mathrm{n}}^2-3=\mathrm{an}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)+\mathrm{bn}(\mathrm{n}-1)(\mathrm{n}-2)+\mathrm{cn}(\mathrm{n}-1)+\mathrm{dn}+\mathrm{t}$$$$\sum _{\mathrm{n}⩾1}({\mathrm{n}}^4+{2\mathrm{n}}^2-3)\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}=\sum _{\mathrm{n}⩾4}\left(\mathrm{a}\frac{{\mathrm{x}}^{\mathrm{n}}}{(\mathrm{n}-4)!}\right)+\mathrm{b}\sum _{\mathrm{n}⩾4}\frac{{\mathrm{x}}^{\mathrm{n}}}{(\mathrm{n}-3))!}+\mathrm{c}\sum _{\mathrm{n}⩾4}\frac{{\mathrm{x}}^{\mathrm{n}}}{(\mathrm{n}-2)!}+\mathrm{d}\sum _{\mathrm{n}⩾4}\frac{{\mathrm{x}}^{\mathrm{n}}}{(\mathrm{n}-1)!}+\mathrm{t}\sum _{\mathrm{n}⩾4}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}$$$$+\{\underset{\mathrm{n}=1}{\sum ^3}({\mathrm{n}}^4+{2\mathrm{n}}^2-3)\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}=\mathrm{h}\left(\mathrm{x}\right)\}$$$$={\mathrm{ax}}^4{\mathrm{e}}^{\mathrm{x}}+{\mathrm{bx}}^3({\mathrm{e}}^{\mathrm{x}}-1)+{\mathrm{cx}}^2({\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x})+\mathrm{dx}({\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}-\frac{{\mathrm{x}}^2}{2})+\mathrm{t}({\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}-\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^3}{6})+\mathrm{h}\left(\mathrm{x}\right)$$$$$$

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