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 :
$(1)$ $\sum_{i=1}^n\ i^2= \frac{n(n+1)(2n+1)}{6}$
$(2)$ $\sum_{i=1}^n\ i^3=\frac{n^2(n+1)^2}{4}$

Let $S(n,t)=\sum_{1 ≤r ≤n} r^t$
We know,
$(r+1)^2-r^2=2\cdot r+1$
Putting r=1,2,3,...,n-1,n we get,
$(1+1)^2-1^2=2\cdot1+1$
$(2+1)^2-2^2=2\cdot2+1$
...
$(n-1+1)^2-(n-1)^2=2\cdot (n-1)+1$
$(n+1)^2-n^2=2\cdot n+1$
Adding both sides, $(n+1)^2-1=2S(n,1)+n$
=>$S(n,1)=\frac{n(n+1)}{2}$
We know, $(r+1)^3-r^3=3\cdot r^2+3\cdot r+1$
Putting r=1,2,3,...,n-1,n and adding them we get,
$(n+1)^3-1=3S(n,2)+3S(n,1)+n$
But $S(n,1)=\frac{n(n+1)}{2}$
$=>3S(n,2)=n^3+3n^2+3n+1-1-n-3\frac{n(n+1)}{2}$
$=>6S(n,2)=2n^3+6n^2+6n-2n-3n(n+1)$
$=2n^3+3n^2+n=n(2n^2+3n+1)=n(n+1)(2n+1)$
$=>S(n,2)=\frac{n(n+1)(2n+1)}{6}$
Now we know, $(r+1)^4-r^4=4\cdot r^3+6\cdot r^2+4\cdot r+1$
Putting r=1,2,3,...,n-1,n and adding them we get,
$(n+1)^4-1=4S(n,3)+6S(n,2)+4S(n,1)+n$
Putting the values $S(n,1)$ and  $S(n,2)$
$n^4+4n^3+6n^2+4n+1-1=4S(n,3)+(2n^3+3n^2+n)+2(n^2+n)+n$
$=>4S(n,3)=n^4+2n^3+n^2$
$=>S(n,3)=(\frac{n(n+1)}{2})^2$
We can go as far as we like.
