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Question Number 61937 by Tawa1 last updated on 12/Jun/19

$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{n}}^3+5}{\mathrm{n}!}$$

Commented by mr W last updated on 12/Jun/19

$$seemstobe10e.$$$$othersshouldprove.$$

Commented by Tawa1 last updated on 12/Jun/19

$$\mathrm{Yes},10\mathrm{e}\mathrm{is}\mathrm{correct},\mathrm{but}\mathrm{workings}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 12/Jun/19

$$letS=\sum _{n=0}^{\mathrm{\infty }}\frac{n^3+5}{n!}\Rightarrow S=\sum _{n=0}^{\mathrm{\infty }}\frac{n^3}{n!}+5\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}$$$$wehave\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}=e^{x}withradiusR=\mathrm{\infty }\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}=e$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{n^3}{n!}=\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{(n-1)!}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(n+1)}^2}{n!}=\sum _{n=0}^{\mathrm{\infty }}\frac{n^2+2n+1}{n!}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{n}{(n-1)!}+2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n-1)!}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{n+1}{n!}+2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}+e=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n-1)!}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}+3e$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}+4e=5e\Rightarrow S=5e+5e=10e.$$$$$$$$$$

Commented by Tawa1 last updated on 12/Jun/19

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

Commented by maxmathsup by imad last updated on 12/Jun/19

$$youarewelcome.$$

Answered by mr W last updated on 12/Jun/19

$$f\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n^3+5)x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n^3{x}^n}{n!}+5\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2{x}^n}{(n-1)!}+5\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(n+1)}^2{x}^{n+1}}{n!}+5\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n^2+2n+1)x^{n+1}}{n!}+5\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n^2{x}^{n+1}}{n!}+2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{nx}^{n+1}}{n!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{n+1}}{n!}+5\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{nx}^{n+1}}{(n-1)!}+2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n+1}}{(n-1)!}+x\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}+5\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n+1)x^{n+2}}{n!}+2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{n+2}}{n!}+(x+5)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n+1)x^{n+2}}{n!}+2x^2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}+(x+5)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n+1)x^{n+2}}{n!}+(2x^2+x+5)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{nx}^{n+2}}{n!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{n+2}}{n!}+(2x^2+x+5)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n+2}}{(n-1)!}+x^2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}+(2x^2+x+5)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{n+3}}{n!}+(3x^2+x+5)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=x^3\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}+(3x^2+x+5)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=(x^3+3x^2+x+5)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$=(x^3+3x^2+x+5)e^x$$$$$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n^3+5)x^n}{n!}=(x^3+3x^2+x+5)e^x$$$$withx=1:$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{n^3+5}{n!}=(1^3+3\times 1^2+1+5)e^1=10e$$$$$$$$withx=2:$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2^n(n^3+5)}{n!}=(2^3+3\times 2^2+2+5)e^2=27e^2$$$$withx=\frac{1}{2}:$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{(n^3+5)}{n!2^n}=(\frac{1}{2^3}+3\times \frac{1}{2^2}+\frac{1}{2}+5)e^{\frac{1}{2}}=\frac{51\sqrt{e}}{8}$$

Commented by mr W last updated on 12/Jun/19

$$thankyousirs!$$

Commented by MJS last updated on 12/Jun/19

$$\mathrm{great}\mathrm{job}!$$

Commented by Tawa1 last updated on 12/Jun/19

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

Commented by Prithwish sen last updated on 12/Jun/19

$$\mathrm{excellent}\mathrm{sir}$$

Commented by malwaan last updated on 13/Jun/19

$$FANTASTIC\mathit{s}\mathit{i}\mathit{r}!$$

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