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.

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

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

**To solve this question we will write in this way**,

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)+1so ,

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

**Therefore,**

** \displaystyle Sn=\frac{{n(n+1)(2n+1)}}{6} [Answer]**

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