Find-the-formular-for-the-sum-of-the-first-kth-power-of-natural-number-

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

do the question mean S_k =1^k +2^k +3^k +...+n^k pls reply

$${do}\:{the}\:{question}\:{mean} \\ $$$${S}_{{k}} =\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +\mathrm{3}^{{k}} +…+{n}^{{k}} \: \\ $$$${pls}\:{reply} \\ $$

Commented by Tawa1 last updated on 06/Oct/18

Yes sir. exactly, Thanks for your help sir. God bless you

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

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

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

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

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

S_r =r∫S_(r−1) dn+nB_r to find S_5 we need the value of S_4 now S_3 =1^3 +2^3 +3^3 +..+n^3 ={((n(n+1))/2)}^2 =((n^2 (n^2 +2n+1))/4)=((n^4 +2n^3 +n^2 )/4) S_4 =4∫((n^4 +2n^3 +n^2 )/4)dn+nB_4 =(n^5 /5)+((2n^4 )/4)+(n^3 /3)+n×((−1)/(30)) =(n^5 /5)+(n^4 /(2 ))+(n^3 /3)−(n/(30)) S_5 =5∫S_4 dn+nB_5 S_5 =5∫(n^5 /5)+(n^4 /2)+(n^3 /3)−(n/(30))dn+nB_5 S_5 =5{(n^6 /(30))+(n^5 /(10))+(n^4 /(12))−(n^2 /(60))}+nB_5 S_5 =(n^6 /6)+(n^5 /2)+((5n^4 )/(12))−(n^2 /(12))+nB_5 value of B_5 =0 so 1^5 +2^5 +3^5 +...+n^5 =(n^6 /6)+(n^5 /2)+((5n^4 )/(12))−(n^2 /(12))

$${S}_{{r}} ={r}\int{S}_{{r}−\mathrm{1}} {dn}+{nB}_{{r}} \\ $$$${to}\:{find}\:{S}_{\mathrm{5}} \:\:{we}\:{need}\:{the}\:{value}\:{of}\:{S}_{\mathrm{4}} \\ $$$${now}\:{S}_{\mathrm{3}} =\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +..+{n}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} =\frac{{n}^{\mathrm{2}} \left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{4}}=\frac{{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{3}} +{n}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${S}_{\mathrm{4}} =\mathrm{4}\int\frac{{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{3}} +{n}^{\mathrm{2}} }{\mathrm{4}}{dn}+{nB}_{\mathrm{4}} \\ $$$$\:\:=\frac{{n}^{\mathrm{5}} }{\mathrm{5}}+\frac{\mathrm{2}{n}^{\mathrm{4}} }{\mathrm{4}}+\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+{n}×\frac{−\mathrm{1}}{\mathrm{30}} \\ $$$$=\frac{{n}^{\mathrm{5}} }{\mathrm{5}}+\frac{{n}^{\mathrm{4}} }{\mathrm{2}\:}+\frac{{n}^{\mathrm{3}} }{\mathrm{3}}−\frac{{n}}{\mathrm{30}} \\ $$$${S}_{\mathrm{5}} =\mathrm{5}\int{S}_{\mathrm{4}} {dn}+{nB}_{\mathrm{5}} \\ $$$${S}_{\mathrm{5}} =\mathrm{5}\int\frac{{n}^{\mathrm{5}} }{\mathrm{5}}+\frac{{n}^{\mathrm{4}} }{\mathrm{2}}+\frac{{n}^{\mathrm{3}} }{\mathrm{3}}−\frac{{n}}{\mathrm{30}}{dn}+{nB}_{\mathrm{5}} \\ $$$${S}_{\mathrm{5}} =\mathrm{5}\left\{\frac{{n}^{\mathrm{6}} }{\mathrm{30}}+\frac{{n}^{\mathrm{5}} }{\mathrm{10}}+\frac{{n}^{\mathrm{4}} }{\mathrm{12}}−\frac{{n}^{\mathrm{2}} }{\mathrm{60}}\right\}+{nB}_{\mathrm{5}} \\ $$$${S}_{\mathrm{5}} =\frac{{n}^{\mathrm{6}} }{\mathrm{6}}+\frac{{n}^{\mathrm{5}} }{\mathrm{2}}+\frac{\mathrm{5}{n}^{\mathrm{4}} }{\mathrm{12}}−\frac{{n}^{\mathrm{2}} }{\mathrm{12}}+{nB}_{\mathrm{5}} \:\: \\ $$$${value}\:{of}\:{B}_{\mathrm{5}} =\mathrm{0}\:{so} \\ $$$$\mathrm{1}^{\mathrm{5}} +\mathrm{2}^{\mathrm{5}} +\mathrm{3}^{\mathrm{5}} +…+{n}^{\mathrm{5}} =\frac{{n}^{\mathrm{6}} }{\mathrm{6}}+\frac{{n}^{\mathrm{5}} }{\mathrm{2}}+\frac{\mathrm{5}{n}^{\mathrm{4}} }{\mathrm{12}}−\frac{{n}^{\mathrm{2}} }{\mathrm{12}} \\ $$$$ \\ $$

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