Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 78941 by M±th+et£s last updated on 21/Jan/20

$$Q.findthesum$$$$S=\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+....$$$$thenfind\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^4}{n!}$$

Answered by mind is power last updated on 21/Jan/20

$$=\sum _{n⩾1}\frac{n^3}{(n-1)!}=\sum _{k⩾0}\frac{{(k+1)}^3}{k!}=\sum _{k⩾0}\frac{(k^3+3k^2+3k+1)}{k!}$$$$=\sum _{k⩾1}\frac{k^2}{(k-1)!}+3\sum _{k⩾1}\frac{k}{(k-1)!}+3\sum _{k⩾1}\frac{1}{(k-1)!}+\sum _{k⩾1}\frac{1}{k!}$$$$=\sum _{k⩾0}\frac{{(k+1)}^2}{k!}+3\sum _{k⩾0}\frac{k+1}{k!}+3\sum _{k⩾0}\frac{1}{k!}+\sum _{k⩾0}\frac{1}{k!}-1$$$$=\sum _{k⩾0}\frac{k^2+2k+1}{k!}+3\sum _{k⩾0}\frac{k}{k!}+\sum _{k⩾0}\frac{7}{k!}-1$$$$=\sum _{k⩾1}\frac{k}{(k-1)!}+2\sum _{k⩾1}\frac{1}{(k-1)!}+\sum _{k⩾0}\frac{1}{k!}+3\sum _{k⩾1}\frac{1}{(k-1)!}+7\sum _{k⩾0}\frac{1}{k!}-1$$$$=\sum _{k⩾0}\frac{k+1}{k!}+2\sum _{k⩾0}\frac{1}{k!}+\sum _{k⩾0}\frac{1}{k!}+10\sum _{k⩾0}\frac{1}{k!}-1$$$$=\sum _{k⩾1}\frac{1}{(k-1)!}+\sum _{k⩾0}\frac{1}{k!}+2\sum _{k⩾0}\frac{1}{k!}+\sum _{k⩾0}\frac{1}{k!}+10\sum _{k⩾0}\frac{1}{k!}-1$$$$=15\sum _{k⩾0}\frac{1}{k!}-1=15e-1$$

Answered by Smail last updated on 22/Jan/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^4}{n!}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^3}{(n-1)(n-2)(n-3)(n-4)!}$$$$\frac{n^3}{(n-1)!}=\frac{1}{(n-1)!}+\frac{7}{(n-2)!}+\frac{6}{(n-3)!}+\frac{1}{(n-4)!}$$$$S=\underset{n=4}{\sum ^{\mathrm{\infty }}}(\frac{1}{(n-1)!}+\frac{7}{(n-2)!}+\frac{6}{(n-3)!}+\frac{1}{(n-4)!})+(1+\frac{16}{2}+\frac{27}{2})$$$$S=\frac{45}{2}+\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{1}{(n-1)!}+\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{7}{(n-2)!}+\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{6}{(n-3)!}+\underset{n=4}{\sum ^{\mathrm{\infty }}}\frac{1}{(n-4)!}$$$$S=\frac{45}{2}+\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}+\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{7}{n!}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{6}{n!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}$$$$=\frac{45}{2}+(e-1-1-\frac{1}{2})+7(e-1-1)+6(e-1)+e$$$$=\frac{45}{2}+15e-\frac{5}{2}-14-6$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n^4}{n!}=15e$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com