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Question Number 44892 by Tawa1 last updated on 06/Oct/18

$$\mathrm{Find}\mathrm{the}\mathrm{formular}\mathrm{for}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{kth}\mathrm{power}\mathrm{of}\mathrm{natural}\mathrm{number}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

$$dothequestionmean$$$$S_k=1^k+2^k+3^k+...+n^k$$$$plsreply$$

Commented by Tawa1 last updated on 06/Oct/18

$$\mathrm{Yes}\mathrm{sir}.\mathrm{exactly},\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{help}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by Tawa1 last updated on 06/Oct/18

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

Commented by Tawa1 last updated on 06/Oct/18

$$\mathrm{Sir},\mathrm{am}\mathrm{trying}\mathrm{to}\mathrm{understand}\mathrm{the}\mathrm{theorem}\mathrm{but}\mathrm{am}\mathrm{confused}.$$$$\mathrm{Can}\mathrm{you}\mathrm{help}\mathrm{me}\mathrm{to}\mathrm{use}\mathrm{the}\mathrm{theorem}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{n}\mathrm{terms}$$$$1^5+2^5+3^5+4^5+...\mathrm{or}\mathrm{any}\mathrm{example}.$$$$\mathrm{Please},\mathrm{i}\mathrm{will}\mathrm{read}\mathrm{through}\mathrm{your}\mathrm{example}\mathrm{and}\mathrm{understand}.\mathrm{God}\mathrm{bless}\mathrm{you}$$$$\mathrm{sir}.\mathrm{And}\mathrm{thanks}\mathrm{for}\mathrm{everytime}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

$$S_r=r\int S_{r-1}dn+{nB}_r$$$$tofind{S}_{5}weneedthevalueof{S}_4$$$$now{S}_3=1^3+2^3+3^3+..+n^3$$$$={\left\{\frac{n(n+1)}{2}\right\}}^2=\frac{n^2(n^2+2n+1)}{4}=\frac{n^4+2n^3+n^2}{4}$$$$S_4=4\int \frac{n^4+2n^3+n^2}{4}dn+{nB}_4$$$$=\frac{n^5}{5}+\frac{2n^4}{4}+\frac{n^3}{3}+n\times \frac{-1}{30}$$$$=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$$$S_5=5\int S_{4}dn+{nB}_5$$$$S_5=5\int \frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}dn+{nB}_5$$$$S_5=5\{\frac{n^6}{30}+\frac{n^5}{10}+\frac{n^4}{12}-\frac{n^2}{60}\}+{nB}_5$$$$S_5=\frac{n^6}{6}+\frac{n^5}{2}+\frac{5n^4}{12}-\frac{n^2}{12}+{nB}_5$$$$valueof{B}_5=0so$$$$1^5+2^5+3^5+...+n^5=\frac{n^6}{6}+\frac{n^5}{2}+\frac{5n^4}{12}-\frac{n^2}{12}$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

Commented by Tawa1 last updated on 06/Oct/18

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}$$

Commented by Tawa1 last updated on 06/Oct/18

$$\mathrm{I}\mathrm{really}\mathrm{understand}\mathrm{now}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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