elementary number theory - how can one find the value of the expression, $(1^2+2^2+3^2+cdots+n^2)$

how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$

Let,

$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$

$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$

$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$

Answer

Hint: $(n+1)^3-n^3=3n^2+3n+1$ and use telescopic sum.