How do we evaluate Double Summation with independant indices? How can we show that this double sum is equal to product of single sum?
$$\sum_{i=0}^{n}\sum_{j=0}^{n}a_ia_j = \left(\sum_{i=0}^{n}a_i\right)^2$$
I tried like this, that for the inner sum, we can take $a_i$ common:
$$\sum_{i=0}^{n}\left[a_i\left(\sum_{j=0}^{n}a_j\right)\right]$$
Then for outer sum we can take $\left(\sum_{j=0}^{n}a_j\right)$ common to get:
$$\left(\sum_{i=0}^{n}a_i\right)\times\left(\sum_{j=0}^{n}a_j\right) = \left(\sum_{i=0}^{n}a_i\right)^2$$
Is this method ok? Also give some insights on double sums and how to evaluate them! Thanks a lot!

hint
$$\sum_{i=0}^n(\sum_{j=0}^na_ia_j )=$$
$$\sum_{i=0}^n(a_i\sum_{j=0}^na_j) $$
$$a_0+a_1+...a_n=\sum_{k=0}^n a_k $$
