Specific double summation problems How should I evaluate double sum of a number? 
Example
$$\sum_{i=0}^n\sum_{k=i+1}^n 4 $$
And how should I evaluate double sum of $$\sum_{i=0}^n\sum_{k=i+1}^n i^2 $$
Note:- I am completely new at these. I am good at single summations though. So if possible explain each and every of the step(I learn quickly). And if possible also provide some reference to further polish my skills.
 A: In this case it's easier to swap order of summation. 
First summation:
$$\sum_{i=0}^n\sum_{k=i+1}^n 4=4\sum_{k=1}^n\sum_{i=0}^{k-1}1=4\sum_{k=1}^nk=2n(n+1)\quad\blacksquare$$
Second summation:
$$\sum_{i=0}^n\sum_{k=i+1}^n i^2=\sum_{k=1}^n\sum_{i=0}^{k-1}i^2=
\sum_{k=1}^n\sum_{i=0}^{k-1}\binom i2+\binom {i+1}2=\sum_{k=n}^n\binom k3+\binom {k+1}3\\
=\binom {n+1}4+\binom {n+2}4=\frac {2n\cdot (n+1)n(n-1)}{4!}=\frac 1{12}n^2(n^2-1)\quad\blacksquare$$
A: Hint:
$$\sum_{i=0}^n\sum_{k=i+1}^n 4 =\sum_{i=0}^n (n-i) 4 = 4n\sum_{i=0}^n 1 - 4\sum_{i=0}^ni  $$
$$\sum_{i=0}^n\sum_{k=i+1}^n i^2 = \sum_{i=0}^n (n- i)i^2 = \sum_{i=0}^n ni^2- i^3 =  n\sum_{i=0}^n i^2- \sum_{i=0}^ni^3$$
A: $$\sum_{i=0}^n\sum_{k=i+1}^n 4 =4 \sum_{i=0}^n (n-i)=4n(n+1)-4\frac{n(n+1)}{2}=2n(n+1)$$
$$\sum_{i=0}^n\sum_{k=i+1}^ni^2=\sum_{i=0}^ni^2(n-i)=n\sum_{i=0}^ni^2-\sum_{i=0}^ni^3$$
$$=n\frac{1}{6}n(n+1)(2n+1)-\frac{1}{4}n^2(n+1)^2$$
$$=\frac{1}{6}n^2(n+1)(2n+1)-\frac{1}{4}n^2(n+1)^2$$
$$=n^2(n+1)\left[\frac{1}{6}(2n+1)-\frac{1}{4}(n+1)\right]$$
$$=\frac{1}{12}n^2(n^2-1)$$
http://mathworld.wolfram.com/PowerSum.html
