Question

$\mathsf{Prove \: that : }$



$\mathsf{\implies \sum \limits_{n \: = \: 1}^{n} {n}^{2} \: = \: \dfrac{n(n \: + \: 1)(2n \: + \: 1)}{6} }$

Answer 1

hey buddy here's ur answer, plz refer attachment...


hope it helps u

Answer 2

We know  that,

$(n-1)^3=n^{3}-3n^{2}+3n-1\\\;\\or,\\\;\\n^{3}-(n-1)^{3}=3n^{2}-3n+1\\\;\\\text{Now, putting n=1,2,3,.......(n-1),n}\\\;\\We,\;get\\\;\\1^{3}-0^{3}=3.1^{2}-3.1+1\\\;\\2^{3}-1^{3}=3.2^{2}-3.2+1\\\;\\3^{3}-2^{3}=3.3^{2}-3.3+1\\\;\\........\;\;\;\;.....\\\;\\.......\;\;\;.........\\\;\\(n-1)^{3}-n^{3}=3(n-1)^{2}-3(n-1)+1\\\;\\n^{3}-(n-1)^{3}=3n^{2}-3n+1\\\;\\\text{Adding them column wise(diagonals will be cancel out and  only\;}n^{3}\text{\;remain in L.H.S}\\\;\\\text{We get,}$

$n^{3}=3(1^{2}+2^3+3^3+4^3+.....+n^2)-3(1+2+3+4+......+n)+(1+1+1+.....\text{upto n terms})\\\;\\n^3=3(\sum{n^2})-3(\sum{n})+n\\\;\\3\sum{n^2}=n^3-n+3\sum{n}\\\;\\3\sum{n^2}=n^3-n+3.\frac{n(n+1)}{2}\\\;\\3\sum{n^2}=n(n^2-1)+3.\frac{n(n+1)}{2}\\\;\\3\sum{n^2}=n(n+1)(n-1)+3.\frac{n(n+1)}{2}\\\;\\3\sum{n^2}=n(n+1)[n-1+\frac{3}{2}]\\\;\\3\sum{n^2}=n(n+1)[\frac{2n-2+3}{2}]\\\;\\3\sum{n^2}=n(n+1)[\frac{2n+1}{2}]\\\;\\3\sum{n^2}=\frac{n(n+1)(2n+1)}{2}\\\;\\\sum{n^2}=\frac{n(n+1)(2n+1)}{6}$

$Note:\\\;\\1+2+3+4+.......n=\frac{n(n+1)}{2}$