Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 108571 by Dwaipayan Shikari last updated on 17/Aug/20

Answered by mr W last updated on 17/Aug/20

$$e^x=1+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!}$$$$e=1+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=e-1$$$$$$$$\frac{e^x-1}{x}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n-1}}{n!}$$$$-\frac{e^x-1}{x^2}+\frac{e^x}{x}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(n-1)\frac{x^{n-2}}{n!}$$$$-\frac{e^x-1}{x}+e^x=\underset{n=1}{\sum ^{\mathrm{\infty }}}(n-1)\frac{x^{n-1}}{n!}$$$$\frac{e^x-1}{x^2}-\frac{e^x}{x}+e^x=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(n-1)}^2\frac{x^{n-2}}{n!}$$$$\frac{e-1}{1}-e+e=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(n-1)}^2}{n!}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(n-1)}^2}{n!}=e-1$$$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2+{(n-1)}^2}{n!}=2(e-1)+(e-1)=3(e-1)$$

Commented by ajfour last updated on 17/Aug/20

$$\mathcal{V}ery\mathcal{N}ice!\mathcal{T}hankyouSir.$$

Commented by Dwaipayan Shikari last updated on 17/Aug/20

$$Greatsir!$$

Answered by Dwaipayan Shikari last updated on 17/Aug/20

$$y_0△y_0{△}^2{y}_0$$$$2$$$$1$$$$32$$$$3$$$$62$$$$5$$$$112$$$$7$$$$18$$$$\varphi \left(y\right)=2+(n-1)+(n-1)(n-2)=2+{(n-1)}^2=n^2-2n+3$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2-2n+3}{n!}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2}{n!}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n}{n!}+\sum ^{\mathrm{\infty }}\frac{3}{n!}$$$$\sum ^{\mathrm{\infty }}\frac{3}{n!}=3(e-1)$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^2}{n!}=\frac{1^2}{1!}+\frac{2^2}{2!}+\frac{3^2}{3!}+....=1+\frac{2}{1!}+\frac{3}{2!}+\frac{4}{3!}+....$$$$=(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...)+(\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+...)$$$$=e+(1+\frac{1}{1!}+\frac{1}{2!}+...)=e+e=2e$$$$\sum ^{\mathrm{\infty }}\frac{2n}{n!}=2(\frac{1}{1!}+\frac{2}{2!}+....)=2(1+\frac{1}{1!}+\frac{1}{2!}+....)=2e$$$$So$$$$\sum ^{\mathrm{\infty }}\frac{n^2-2n+3}{n!}=3(e-1)$$

Commented by Dwaipayan Shikari last updated on 17/Aug/20

$$Thisismyapproach$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com