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Question Number 45585 by pieroo last updated on 14/Oct/18
prove that1^3 +2^3 +3^3 +...+n^3 =((n^2 (n+1)^2 )/4) for every natural number n
$$\mathrm{prove}\:\mathrm{that1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+\mathrm{n}^{\mathrm{3}} =\frac{\mathrm{n}^{\mathrm{2}} \left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{natural}\:\mathrm{number}\:\boldsymbol{\mathrm{n}} \\ $$
Commented by maxmathsup by imad last updated on 14/Oct/18
let p(x)=((x^2 (x−1)^2 )/4) we have p(x+1)−p(x)=(((x+1)^2 x^2 )/4) −((x^2 (x−1)^2 )/4) =(x^2 /4){(x+1)^2 −(x−1)^2 }=(x^2 /4){x^2 +2x+1−x^2 +2x−1}=x^3 ⇒ Σ_(k=1) ^n (p(k+1)−p(k))=Σ_(k=1) ^n k^3 ⇒Σ_(k=1) ^n k^3 =p(n+1)−p(1 =(((n+1)^2 n^2 )/4) =((n^2 (n+1)^2 )/4) .
$${let}\:{p}\left({x}\right)=\frac{{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\:{we}\:{have}\:{p}\left({x}+\mathrm{1}\right)−{p}\left({x}\right)=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{4}}\:−\frac{{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right\}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\left\{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\right\}={x}^{\mathrm{3}} \:\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \left({p}\left({k}+\mathrm{1}\right)−{p}\left({k}\right)\right)=\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{3}} \:\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{3}} \:={p}\left({n}+\mathrm{1}\right)−{p}\left(\mathrm{1}\right. \\ $$$$=\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} {n}^{\mathrm{2}} }{\mathrm{4}}\:\:\:=\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:. \\ $$
Commented by pieroo last updated on 14/Oct/18
thanks very much sir.
$$\mathrm{thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
another general method... S_r =1^r +2^r +3^r +...+n^r formuls is S_r =r∫S_(r−1) dn+nB_r now if we know s_1 =1+2+3+...+n we can find S_2 =1^2 +2^2 +3^2 +...+n^2 S_3 =1^3 +2^3 +3^3 +..+n^3 S_(40) =1^(40) +2^(40) +3^(40) +...+n^(40) now s_1 =((n(n+1))/2)=(n^2 /2)+(n/2) S_r =r∫S_(r−1) dn+nB_r so S_2 =2∫S_1 dn+nB_2 =2∫((n^2 /2)+(n/2))dn+nB_2 =2[(n^3 /6)+(n^2 /4)]+n×(1/6) B_2 =(1/6) (see table) =(n^3 /3)+(n^2 /2)+(n/6)=((2n^3 +3n^2 +n)/6) =((n(2n^2 +3n+1))/6)=(n/6)(2n^2 +2n+n+1) =((n(n+1)(2n+1))/6) using this we can find S_3 =3∫S_2 dn+nB_3 S_3 =3∫(n^3 /3)+(n^2 /2)+(n/6)+n×0 B_3 =0 =3((n^4 /(12))+(n^3 /6)+(n^2 /(12)))=(n^4 /4)+(n^3 /2)+(n^2 /4)=(n^2 /4)(n^2 +2n+1) =(n^2 /4)(n+1)^2 ={((n(n+1))/2)}^2 is the answer
$${another}\:{general}\:{method}… \\ $$$${S}_{{r}} =\mathrm{1}^{{r}} +\mathrm{2}^{{r}} +\mathrm{3}^{{r}} +…+{n}^{{r}} \\ $$$${formuls}\:{is}\:{S}_{{r}} ={r}\int{S}_{{r}−\mathrm{1}} {dn}+{nB}_{{r}} \\ $$$${now}\:{if}\:{we}\:{know}\:{s}_{\mathrm{1}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n} \\ $$$${we}\:{can}\:{find}\:{S}_{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} \\ $$$${S}_{\mathrm{3}} =\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +..+{n}^{\mathrm{3}} \\ $$$${S}_{\mathrm{40}} =\mathrm{1}^{\mathrm{40}} +\mathrm{2}^{\mathrm{40}} +\mathrm{3}^{\mathrm{40}} +…+{n}^{\mathrm{40}} \\ $$$${now}\:{s}_{\mathrm{1}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{2}} \\ $$$${S}_{{r}} ={r}\int{S}_{{r}−\mathrm{1}} {dn}+{nB}_{{r}} \\ $$$${so}\:{S}_{\mathrm{2}} =\mathrm{2}\int{S}_{\mathrm{1}} {dn}+{nB}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int\left(\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{2}}\right){dn}+{nB}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left[\frac{{n}^{\mathrm{3}} }{\mathrm{6}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\right]+{n}×\frac{\mathrm{1}}{\mathrm{6}}\:\:\:{B}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\:\left({see}\:{table}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}}=\frac{\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:=\frac{{n}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right)}{\mathrm{6}}=\frac{{n}}{\mathrm{6}}\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}+{n}+\mathrm{1}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${using}\:{this}\:{we}\:{can}\:{find}\:{S}_{\mathrm{3}} =\mathrm{3}\int{S}_{\mathrm{2}} {dn}+{nB}_{\mathrm{3}} \\ $$$${S}_{\mathrm{3}} =\mathrm{3}\int\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}}+{n}×\mathrm{0}\:\:\:\:{B}_{\mathrm{3}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:=\mathrm{3}\left(\frac{{n}^{\mathrm{4}} }{\mathrm{12}}+\frac{{n}^{\mathrm{3}} }{\mathrm{6}}+\frac{{n}^{\mathrm{2}} }{\mathrm{12}}\right)=\frac{{n}^{\mathrm{4}} }{\mathrm{4}}+\frac{{n}^{\mathrm{3}} }{\mathrm{2}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}=\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\:\:\:\:=\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\left({n}+\mathrm{1}\right)^{\mathrm{2}} =\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} \:\:\:{is}\:{the}\:{answer} \\ $$
Commented by pieroo last updated on 14/Oct/18
wow, i love this
$$\mathrm{wow},\:\mathrm{i}\:\mathrm{love}\:\mathrm{this}\: \\ $$
Commented by pieroo last updated on 14/Oct/18
Sir thank you very much, but please help me with Question number 45353
$$\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much},\:\mathrm{but}\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with} \\ $$$$\mathrm{Question}\:\mathrm{number}\:\mathrm{45353} \\ $$

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