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Question Number 114110 by bemath last updated on 17/Sep/20

$$\frac{4}{1!}+\frac{11}{2!}+\frac{22}{3!}+\frac{37}{4!}+...=?$$

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

$$y△y{△}^{2}y$$$$4$$$$7$$$$114$$$$11$$$$224$$$$15$$$$37$$$$\varphi \left(y\right)=4+7(n-1)+2(n-1)(n-2)$$$$=4+(n-1)(7+2n-4)$$$$=4+(n-1)(2n+3)=4+2n^2+n-3=2n^2+n+1$$$$\sum ^{\mathrm{\infty }}\frac{2n^2+n+1}{n!}=4e+e+e-1=6e-1$$$$\sum ^{\mathrm{\infty }}\frac{2n^2}{n!}=2(\frac{1^2}{1!}+\frac{2^2}{2!}+\frac{3^2}{3!}+...)=2(1+\frac{2}{1!}+\frac{3}{2!}+....)$$$$=2((\frac{1}{1!}+\frac{2}{2!}+..)+(1+\frac{1}{2!}+...))=2.2e=4e$$$$\sum ^{\mathrm{\infty }}\frac{n}{n!}=\frac{1}{1!}+\frac{2}{2!}+.....=(1+\frac{1}{1!}+\frac{1}{2!}+....)=e$$$$\sum ^{\mathrm{\infty }}\frac{1}{n!}=e-1$$

Commented by bemath last updated on 17/Sep/20

$$bravoo$$

Answered by bobhans last updated on 17/Sep/20

$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n^2+n+1}{n!}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n(n-1)+3n+1}{n!}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n(n-1)}{n!}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{3n}{n!}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}$$$$S_1=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=e-1.[e^x=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n!},lettingx=1]$$$$S_2=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{3n}{n!}=3e.[e^x=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{nx}^{n-1}}{n!}]$$$$S_3=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n(n-1)}{n!}=2e.[e^x=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n(n-1)x^{n-2}}{n!}]$$$$HenceS=S_1+S_2+S_3=e-1+3e+2e=6e-1$$

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