# How to solve 1^2+2^2+3^2 to n^2 formula

Welcome to my article How to solve 1^2+2^2+3^2 to n^2 formula. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve 1^2+2^2+3^2 to n^2 formula, read and understand it carefully till the end.

Let us know how to solve this question How to solve 1^2+2^2+3^2 to n^2 formula.

First write the question on the page of the notebook.

## How to solve 1^2+2^2+3^2 to n^2 formula

To solve this question we will write in this way,

\displaystyle {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+……….+{{n}^{2}}

\displaystyle Let~Sn={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+…..+{{n}^{2}}

\displaystyle Putting~k=1,2,….,n~successively,\text{ }we\text{ }obtain

consider the identity \displaystyle {{k}^{3}}-{{(k-1)}^{3}}=3{{k}^{2}}-3k+1

putting k=1,2,3………….n successively we obtain .

\displaystyle {{1}^{3}}-{{(1-1)}^{3}}=3{{(1)}^{2}}-3(1)+1

\displaystyle {{2}^{3}}-{{(2-1)}^{3}}=3{{(2)}^{2}}-3(2)+1

so ,

\displaystyle {{n}^{3}}-{{(n-1)}^{3}}=3{{(n)}^{2}}-3(n)+1

Adding both sides we get

\displaystyle {{n}^{3}}-{{(0)}^{3}}=3({{1}^{2}}+{{2}^{2}}+{{3}^{2}}……….{{n}^{2}})-3(1+2+3…….+n)+n

\displaystyle {{n}^{3}}=3\sum\nolimits_{{k=1}}^{n}{{{{k}^{2}}}}-3 \displaystyle \sum\nolimits_{{n=1}}^{n}{{k+n}}

Since ,

\displaystyle \sum\nolimits_{{k=1}}^{n}{{k=1+2+3+4…….+n}}=\frac{{n(n+1)}}{2}

Therefore,

\displaystyle Sn=\sum\nolimits_{{k=1}}^{n}{{{{k}^{2}}=\frac{1}{3}}}\left[ {{{n}^{3}}+\frac{{3n(n+1)}}{2}-n} \right]

\displaystyle Sn=\sum\nolimits_{{k=1}}^{n}{{{{k}^{2}}=\frac{1}{6}}}\left[ {2{{n}^{3}}+3{{n}^{2}}+n} \right]

\displaystyle Sn=\frac{{n(n+1)(2n+1)}}{6}