Finite summation 
Possible Duplicate:
why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$ 

What is the proof without induction for :
$(1)$ $\sum_{i=1}^n\ i^2= \frac{n(n+1)(2n+1)}{6}$
$(2)$ $\sum_{i=1}^n\ i^3=\frac{n^2(n+1)^2}{4}$

There are probably many ways. For example, if you already know that $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, it follows $$3\sum_{i=1}^n i^2 + 3\sum_{i=1}^n i + n = \sum_{i=1}^n (3i^2 + 3i + 1) = \sum_{i=1}^n ((i+1)^3 - i^3) = (n+1)^3 - 1$$ since the last sum telescopes, so $$\sum_{i=1}^n i^2 = \frac{1}{3}((n+1)^3 - 1 - 3\sum_{i=1}^n i - n) = \frac{1}{3}(n^3 + 3n^2 + 2n - \frac{3}{2}n(n+1))$$ $$=\frac{1}{6}(2n^3 + 3n^2 + n) = \frac{n(n+1)(2n+1)}{6}.$$
The second problem can be done in the same way. 
