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Question Number 222249 by MrGaster last updated on 21/Jun/25
Prove:∀n∈Z^+ ,1^3 +2^3 +…+n^3 =(1+2+…+n)^2
$$\mathrm{Prove}:\forall{n}\in\mathbb{Z}^{+} ,\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\ldots+{n}^{\mathrm{3}} =\left(\mathrm{1}+\mathrm{2}+\ldots+{n}\right)^{\mathrm{2}} \\ $$
Answered by MrGaster last updated on 21/Jun/25
Answered by vnm last updated on 21/Jun/25
(Σ_(k=1) ^n k)^2 =((Σ_(k=1) ^(n−1) k)+n)^2 = (Σ_(k=1) ^(n−1) k)^2 +2n(Σ_(k=1) ^(n−1) k)+n^2 = by induction Σ_(k=1) ^(n−1) k^3 +2n((n(n−1))/2)+n^2 = Σ_(k=1) ^(n−1) k^3 +n^2 ((n−1)+1)=Σ_(k=1) ^n k^3
$$\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\right)^{\mathrm{2}} =\left(\left(\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k}\right)+{n}\right)^{\mathrm{2}} = \\ $$$$\left(\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k}\right)^{\mathrm{2}} +\mathrm{2}{n}\left(\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k}\right)+{n}^{\mathrm{2}} = \\ $$$$\mathrm{by}\:\mathrm{induction} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k}^{\mathrm{3}} +\mathrm{2}{n}\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+{n}^{\mathrm{2}} = \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k}^{\mathrm{3}} +{n}^{\mathrm{2}} \left(\left({n}−\mathrm{1}\right)+\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} \\ $$
Answered by mr W last updated on 21/Jun/25
(k+1)^4 −k^4 =4k^3 +6k^2 +4k+1 Σ_(k=1) ^n [(k+1)^4 −k^4 ]=4Σ_(k=1) ^n k^3 +6Σ_(k=1) ^n k^2 +4Σ_(k=1) ^n k+Σ_(k=1) ^n 1 (n+1)^4 −1^4 =4Σ_(k=1) ^n k^3 +6×((n(n+1)(2n+1))/6)+4×((n(n+1))/2)+n 4Σ_(k=1) ^n k^3 =(n+1)[(n+1)^3 −n(2n+3)−1] 4Σ_(k=1) ^n k^3 =(n+1)[n^3 +3n^2 +3n+1−2n^2 −3n−1] 4Σ_(k=1) ^n k^3 =n^2 (n+1)^2 ⇒Σ_(k=1) ^n k^3 =[((n(n+1))/2)]^2 =[Σ_(k=1) ^n k]^2 ✓
$$\left({k}+\mathrm{1}\right)^{\mathrm{4}} −{k}^{\mathrm{4}} =\mathrm{4}{k}^{\mathrm{3}} +\mathrm{6}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left({k}+\mathrm{1}\right)^{\mathrm{4}} −{k}^{\mathrm{4}} \right]=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} +\mathrm{6}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} +\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1} \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} =\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} +\mathrm{6}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\mathrm{4}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{n} \\ $$$$\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} =\left({n}+\mathrm{1}\right)\left[\left({n}+\mathrm{1}\right)^{\mathrm{3}} −{n}\left(\mathrm{2}{n}+\mathrm{3}\right)−\mathrm{1}\right] \\ $$$$\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} =\left({n}+\mathrm{1}\right)\left[{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}−\mathrm{2}{n}^{\mathrm{2}} −\mathrm{3}{n}−\mathrm{1}\right] \\ $$$$\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} ={n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} =\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} =\left[\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\right]^{\mathrm{2}} \:\checkmark \\ $$
Commented by mr W last updated on 21/Jun/25
generally say S_r =Σ_(k=1) ^n k^r (k+1)^(r+1) −k^(r+1) =Σ_(m=0) ^r (((r+1)),(m) )k^m = (((r+1)),(r) )k^r +Σ_(m=0) ^(r−1) (((r+1)),(m) )k^m (n+1)^(r+1) −1= (((r+1)),(r) )Σ_(k=1) ^n k^r +Σ_(m=0) ^(r−1) (((r+1)),(m) )Σ_(k=1) ^n k^m (n+1)^(r+1) −1=(r+1)S_r +Σ_(m=1) ^(r−1) (((r+1)),(m) )S_m +n ⇒S_r =(((n+1)^(r+1) −n−1−Σ_(m=1) ^(r−1) (((r+1)),(m) )S_m )/(r+1)) we can get in this way S_2 from S_1 and S_3 from S_1 , S_2 and S_4 from S_1 , S_2 , S_3 etc.
$${generally}\: \\ $$$${say}\:{S}_{{r}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{r}} \\ $$$$\left({k}+\mathrm{1}\right)^{{r}+\mathrm{1}} −{k}^{{r}+\mathrm{1}} =\underset{{m}=\mathrm{0}} {\overset{{r}} {\sum}}\begin{pmatrix}{{r}+\mathrm{1}}\\{{m}}\end{pmatrix}{k}^{{m}} =\begin{pmatrix}{{r}+\mathrm{1}}\\{{r}}\end{pmatrix}{k}^{{r}} +\underset{{m}=\mathrm{0}} {\overset{{r}−\mathrm{1}} {\sum}}\begin{pmatrix}{{r}+\mathrm{1}}\\{{m}}\end{pmatrix}{k}^{{m}} \\ $$$$\left({n}+\mathrm{1}\right)^{{r}+\mathrm{1}} −\mathrm{1}=\begin{pmatrix}{{r}+\mathrm{1}}\\{{r}}\end{pmatrix}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{r}} +\underset{{m}=\mathrm{0}} {\overset{{r}−\mathrm{1}} {\sum}}\begin{pmatrix}{{r}+\mathrm{1}}\\{{m}}\end{pmatrix}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{m}} \\ $$$$\left({n}+\mathrm{1}\right)^{{r}+\mathrm{1}} −\mathrm{1}=\left({r}+\mathrm{1}\right){S}_{{r}} +\underset{{m}=\mathrm{1}} {\overset{{r}−\mathrm{1}} {\sum}}\begin{pmatrix}{{r}+\mathrm{1}}\\{{m}}\end{pmatrix}{S}_{{m}} +{n} \\ $$$$\Rightarrow{S}_{{r}} =\frac{\left({n}+\mathrm{1}\right)^{{r}+\mathrm{1}} −{n}−\mathrm{1}−\underset{{m}=\mathrm{1}} {\overset{{r}−\mathrm{1}} {\sum}}\begin{pmatrix}{{r}+\mathrm{1}}\\{{m}}\end{pmatrix}{S}_{{m}} }{{r}+\mathrm{1}} \\ $$$${we}\:{can}\:{get}\:{in}\:{this}\:{way}\:{S}_{\mathrm{2}} \:{from}\:{S}_{\mathrm{1}} \:{and} \\ $$$${S}_{\mathrm{3}} \:{from}\:{S}_{\mathrm{1}} ,\:{S}_{\mathrm{2}} \:{and}\:{S}_{\mathrm{4}} \:{from}\:\:{S}_{\mathrm{1}} ,\:{S}_{\mathrm{2}} ,\:{S}_{\mathrm{3}} \\ $$$${etc}. \\ $$
Commented by mr W last updated on 21/Jun/25

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