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.
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: $$\sum_{i=0}^n\sum_{k=i+1}^n i^2=\sum_{k=1}^n\sum_{i=0}^{k-1}i^2= \sum_{k=1}^n\sum_{i=0}^{k-1}\binom i2+\binom {i+1}2=\sum_{k=n}^n\binom k3+\binom {k+1}3\\ =\binom {n+1}4+\binom {n+2}4=\frac {2n\cdot (n+1)n(n-1)}{4!}=\frac 1{12}n^2(n^2-1)\quad\blacksquare$$
Hint: $$\sum_{i=0}^n\sum_{k=i+1}^n 4 =\sum_{i=0}^n (n-i) 4 = 4n\sum_{i=0}^n 1 - 4\sum_{i=0}^ni $$ $$\sum_{i=0}^n\sum_{k=i+1}^n i^2 = \sum_{i=0}^n (n- i)i^2 = \sum_{i=0}^n ni^2- i^3 = n\sum_{i=0}^n i^2- \sum_{i=0}^ni^3$$
$$\sum_{i=0}^n\sum_{k=i+1}^n 4 =4 \sum_{i=0}^n (n-i)=4n(n+1)-4\frac{n(n+1)}{2}=2n(n+1)$$
$$\sum_{i=0}^n\sum_{k=i+1}^ni^2=\sum_{i=0}^ni^2(n-i)=n\sum_{i=0}^ni^2-\sum_{i=0}^ni^3$$ $$=n\frac{1}{6}n(n+1)(2n+1)-\frac{1}{4}n^2(n+1)^2$$ $$=\frac{1}{6}n^2(n+1)(2n+1)-\frac{1}{4}n^2(n+1)^2$$ $$=n^2(n+1)\left[\frac{1}{6}(2n+1)-\frac{1}{4}(n+1)\right]$$ $$=\frac{1}{12}n^2(n^2-1)$$
