Question Number 96527 by student work last updated on 02/Jun/20
$$\mathrm{proof}\:\mathrm{that}\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +....+\mathrm{n}^{\mathrm{2}} =\frac{\mathrm{n}\left(\mathrm{2n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$
Commented by 06122004 last updated on 02/Jun/20
Answered by Farruxjano last updated on 02/Jun/20
Commented by student work last updated on 02/Jun/20
$$\mathrm{thanks}\:\mathrm{dear}\:\mathrm{great} \\ $$
Answered by Sourav mridha last updated on 02/Jun/20
![there are many useful techniques but I want to show very unusual methond... using Euler −Maclaurin formula Σ_(m=1) ^n f(m)=∫_1 ^n f(x)dx+(1/2)f(1)+(1/2)f(n) +(B_2 /(2!))[f^′ (n)−f^′ (1)]+........... where B_n =Bernoulli number and here B_2 =(1/6)...f(m)=m^2 ...so f(1)=1 and f(n)=n^2 f(x)=x^2 so f^′ (n)=2n,f^′ (1)=2 soΣ_(m=1) ^n m^2 =(n^3 /3)−(1/3)+(1/2)+(n^2 /2)+(((n−1))/6) =((2n^3 +3n^2 +n)/6) =((n[2n(n+1)+1(n+1)])/6) =(n/6).(n+1)(2n+1).](Q96535.png)
$$\boldsymbol{{there}}\:\boldsymbol{{are}}\:\mathrm{many}\:\boldsymbol{{useful}}\:\boldsymbol{{technique}}\mathrm{s} \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{I}}\:\boldsymbol{{want}}\:\boldsymbol{{to}}\:\boldsymbol{{show}}\:\boldsymbol{{very}}\: \\ $$$$\boldsymbol{{unusual}}\:\boldsymbol{{methond}}... \\ $$$$\boldsymbol{{using}}\:\boldsymbol{{Euler}}\:−\boldsymbol{{Maclaurin}}\:\boldsymbol{\mathrm{formula}} \\ $$$$\:\: \\ $$$$\underset{\boldsymbol{{m}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{f}}\left(\boldsymbol{{m}}\right)=\int_{\mathrm{1}} ^{\boldsymbol{{n}}} \boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{f}}\left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\boldsymbol{\mathrm{B}}_{\mathrm{2}} }{\mathrm{2}!}\left[\boldsymbol{{f}}^{'} \left(\boldsymbol{{n}}\right)−\boldsymbol{{f}}^{'} \left(\mathrm{1}\right)\right]+........... \\ $$$$\boldsymbol{{where}}\:\boldsymbol{{B}}_{\boldsymbol{{n}}} =\boldsymbol{{Bernoulli}}\:\boldsymbol{{number}} \\ $$$$\boldsymbol{{and}}\:\mathrm{here}\:\:\mathrm{B}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}...\boldsymbol{{f}}\left(\boldsymbol{{m}}\right)=\boldsymbol{{m}}^{\mathrm{2}} ...\boldsymbol{{so}} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1}\:\boldsymbol{{and}}\:\boldsymbol{{f}}\left(\boldsymbol{{n}}\right)=\boldsymbol{{n}}^{\mathrm{2}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\mathrm{2}} \:\boldsymbol{{so}}\:\boldsymbol{{f}}^{'} \left(\boldsymbol{{n}}\right)=\mathrm{2}\boldsymbol{{n}},\boldsymbol{{f}}^{'} \left(\mathrm{1}\right)=\mathrm{2} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{so}}\underset{\boldsymbol{{m}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{m}}^{\mathrm{2}} =\frac{\boldsymbol{{n}}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\boldsymbol{{n}}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\boldsymbol{{n}}−\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\boldsymbol{{n}}^{\mathrm{3}} +\mathrm{3}\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{n}}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\boldsymbol{{n}}\left[\mathrm{2}\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{1}\right)+\mathrm{1}\left(\boldsymbol{{n}}+\mathrm{1}\right)\right]}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\boldsymbol{{n}}}{\mathrm{6}}.\left(\boldsymbol{{n}}+\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right).\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$