$\sum_{n=0}^3 n = 6$ and $6^2 = 36$
I know the formula is $$\left( \sum_{n=0}^\infty a_n \right)\left( \sum_{n=0}^\infty b_n \right) = \sum_{n=0}^\infty \sum_{k=0}^n (a_n)(b_{n-k})$$
So: $$\left( \sum_{n=0}^\infty a_n \right)^2 = \sum_{n=0}^\infty \sum_{k=0}^n (a_n)(a_{n-k})$$
But the formula shows a different story, like: $(a_0)(a_0) + (a_1)(a_1) + (a_1)(a_0) + (a_2)(a_2) + \ldots$ equals 0(0) + 1(1) + 1(0) + 2(2) + 2(1) + 2(0) + 3(3) + 3(2) + 3(1) + 3(0) still doesn’t equal 36 it equals 25, so am I getting the formula wrong or what?
It seems that your calculator is doing $$\sum_{n=0}^3 n^2,$$ not $$\left(\sum_{n=0}^3 n\right)^2.$$
You seem to rely on your calculator too much, and lack appropriate calculating abilities. You should have realized that $14$ is $1+4+9$.