Question

Prove that $1^2+2^2+3^2+.......x^2=\frac{x\left(x+1\right)\left(2x+1\right)}{6}$

Answer 1

Solution :-

We have to find sum of square of first n natural

number

$1^2+2^2+3^2+...+x^2=\sum _{x=1}^x{n}^2$

we know $(x-1{)}^3=x^3-3x^2+3x-1$

$\therefore x^3-(x-1{)}^3=3x^2-3x+1$

Now putting $x=1,2,3,...n,$ we get

$(1{)}^3-(0{)}^3=3(1{)}^2-3(1)+1$

$(2{)}^3-(1{)}^3=3(2{)}^2-3(2)+1$

$(3{)}^3-(2{)}^3=3(3{)}^2-3(3)+1$

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

Adding respective terms of there n equation, we get

$(n{)}^3-(0{)}^3=3(1^2+2^2+3^2+...n^2)-3(1+2+...n)+(1+1+....nterms)$

$n^3=3(1^2+2^2+....+n^2)-3\sum _{n=1}^{n}n+n$

$n^3=3\sum _{n=1}^n{n}^2-3\sum _{n=1}^{n}n+n$

$n^3=3\sum _{n=1}^n{n}^2-3\sum _{n=1}^{n}n+n$

$n^3+3\times \frac{n(n+1)}{2}-n=3\sum _{n=1}^n{n}^2$

$\Rightarrow \sum n^2=\frac{n(n+1)(2n+1)}{6}$