Question Number 103796 by bemath last updated on 17/Jul/20
$$\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{5}} \:=\:? \\ $$
Commented by mr W last updated on 17/Jul/20
Commented by bemath last updated on 17/Jul/20
$${waw}..{complete}\:{sir}. \\ $$$${haha}==.{cooll}\: \\ $$
Answered by Ar Brandon last updated on 17/Jul/20
![Consider; k^6 −(k−1)^6 =k^6 −(k^6 −6k^5 +15k^4 −20k^3 +15k^2 −6k+1) =6k^5 −15k^4 +20k^3 −15k^2 +6k−1 ⇒Σ_(k=1) ^n [k^6 −(k−1)^6 ]=6Σ_(k=1) ^n k^5 +Σ_(k=1) ^n [−15k^4 +20k^3 −15k^2 +6k−1] ⇒n^6 =6Σ_(k=1) ^n k^5 −15Σ_(k=1) ^n k^4 +20Σ_(k=1) ^n k^3 −15Σ_(k=1) ^n k^2 +6Σ_(k=1) ^n k−Σ_(k=1) ^n (1) Substituting the previous series and making Σ_(k=1) ^n k^5 the subject, we get our answer.](Q103812.png)
$$\mathrm{Consider}; \\ $$$$\mathrm{k}^{\mathrm{6}} −\left(\mathrm{k}−\mathrm{1}\right)^{\mathrm{6}} =\mathrm{k}^{\mathrm{6}} −\left(\mathrm{k}^{\mathrm{6}} −\mathrm{6k}^{\mathrm{5}} +\mathrm{15k}^{\mathrm{4}} −\mathrm{20k}^{\mathrm{3}} +\mathrm{15k}^{\mathrm{2}} −\mathrm{6k}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6k}^{\mathrm{5}} −\mathrm{15k}^{\mathrm{4}} +\mathrm{20k}^{\mathrm{3}} −\mathrm{15k}^{\mathrm{2}} +\mathrm{6k}−\mathrm{1} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left[\mathrm{k}^{\mathrm{6}} −\left(\mathrm{k}−\mathrm{1}\right)^{\mathrm{6}} \right]=\mathrm{6}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{5}} +\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left[−\mathrm{15k}^{\mathrm{4}} +\mathrm{20k}^{\mathrm{3}} −\mathrm{15k}^{\mathrm{2}} +\mathrm{6k}−\mathrm{1}\right] \\ $$$$\Rightarrow\mathrm{n}^{\mathrm{6}} =\mathrm{6}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{5}} −\mathrm{15}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{4}} +\mathrm{20}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{3}} −\mathrm{15}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{2}} +\mathrm{6}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}−\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{1}\right) \\ $$$${Substituting}\:{the}\:{previous}\:{series}\:{and}\:{making}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}^{\mathrm{5}} \\ $$$${the}\:{subject},\:{we}\:{get}\:{our}\:{answer}. \\ $$