how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$ 
Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers? 

how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Let,
$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$
$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$
$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$

The claim is that 
$\sum_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$
We will verify this by induction.
Clearly $n = 1$ holds. 
Suppose the formula holds for $n$. Lets verify it holds for $n + 1$. 
$$\sum_{i = 1}^{n + 1} i^2 = \sum_{i = 1}^n i^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2
\\
= \frac{n(n + 1)(2n + 1)}{6} + \frac{6(n^2 + 2n + 1)}{6}
\\
= \frac{2n^3 + 3n^2 + n}{6} + \frac{6n^2 + 12 n + 6}{6}
\\
= \frac{2n^3 + 9n^2 + 13n + 6}{6}$$
If you factor you get
$$= \frac{(n + 1)(n + 2)(2n + 3)}{6} = \frac{(n + 1)((n + 1) + 1)(2(n + 1) + 1)}{6}$$
The result follows for $n + 1$. So by induction the formula holds. 
