Specific double summation problems

Question

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.

Answer 1

$$\sum_{i=0}^n\sum_{k=i+1}^n 4 =4 \sum_{i=0}^n (n-i)=4n(n+1)-4\frac{n(n+1)}{2}=2n(n+1)$$

$$\sum_{i=0}^n\sum_{k=i+1}^ni^2=\sum_{i=0}^ni^2(n-i)=n\sum_{i=0}^ni^2-\sum_{i=0}^ni^3$$ $$=n\frac{1}{6}n(n+1)(2n+1)-\frac{1}{4}n^2(n+1)^2$$ $$=\frac{1}{6}n^2(n+1)(2n+1)-\frac{1}{4}n^2(n+1)^2$$ $$=n^2(n+1)\left[\frac{1}{6}(2n+1)-\frac{1}{4}(n+1)\right]$$ $$=\frac{1}{12}n^2(n^2-1)$$

http://mathworld.wolfram.com/PowerSum.html

Answer 2

Hint: $$\sum_{i=0}^n\sum_{k=i+1}^n 4 =\sum_{i=0}^n (n-i) 4 = 4n\sum_{i=0}^n 1 - 4\sum_{i=0}^ni $$ $$\sum_{i=0}^n\sum_{k=i+1}^n i^2 = \sum_{i=0}^n (n- i)i^2 = \sum_{i=0}^n ni^2- i^3 = n\sum_{i=0}^n i^2- \sum_{i=0}^ni^3$$

Answer 3

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: $$\sum_{i=0}^n\sum_{k=i+1}^n i^2=\sum_{k=1}^n\sum_{i=0}^{k-1}i^2= \sum_{k=1}^n\sum_{i=0}^{k-1}\binom i2+\binom {i+1}2=\sum_{k=n}^n\binom k3+\binom {k+1}3\\ =\binom {n+1}4+\binom {n+2}4=\frac {2n\cdot (n+1)n(n-1)}{4!}=\frac 1{12}n^2(n^2-1)\quad\blacksquare$$