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,
$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$
$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$
$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$
 A: In general, $$\sum_{i=1}^{n}i^{2} = \frac{n(n+1)(2n+1)}{6}.$$
A collection of proofs of this fact can be found here.
A: Hint: $(n+1)^3-n^3=3n^2+3n+1$ and use telescopic sum.
A: 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. 
A: If you can find or sketch some 3D blocks, there is a fun geometric proof. 
Fix some $n$. If you are doing this with real blocks, $n=3$ or $4$ should be convincing. Let's take $n=4$ for now.
Make a $4\times 4\times 1$ base, laid flat, which of course has volume $4^2$.
Now make a $3\times 3 \times 1$ brick and place it, laid flat, above the base with some corners aligned. Continue in this way up to the top, making smaller squares and always aligning with the same corner. You now have a 3D corner of stairs whose volume is $1^2+\cdots +n^2$.
Now the fun part. Make 5 more of these "stair corners", for a total of 6. These six toys can be turned sideways and upside down, and then pieced together to make an $n\times(n+1)\times(2n+1)$ rectangular solid.
