Simplify double summation I need to simplify this to find the Big-Oh but I am not very familiar with double summations (second element is my attempt but it might be wrong):
$$\sum\limits_{i=1}^n \sum\limits_{j=2}^{2n+1}1 = 2\sum\limits_{i=1}^nn$$ Is this correct?

$$\sum_{m=t}^{n}\sum_{q=a}^{cn+d}l=\sum_{m=t}^{n}\left[\sum_{q=a}^{cn+d}l\right]=\sum_{m=t}^{n}\left[l(1+d+cn-a)\right]=l(n-t+1)(1+d+cn-a)$$
So, in your example $t=1,a=2,d=1,c=2,l=1$:
$$\sum_{m=1}^{n}\sum_{q=2}^{2n+1}1=\sum_{m=1}^{n}\left[\sum_{q=2}^{2n+1}1\right]=\sum_{m=1}^{n}\left[1(1+1+2n-2)\right]=$$
$$1(n-1+1)(1+1+2n-2)=2n^2$$
