Find sum of $\sum_{i=1}^{8}$ $\sum_{j=1}^{10}$ $(2i+2j)$

Question

The question is how to find $\sum_{i=1}^{8}$ $\sum_{j=1}^{10}$ $(2i+2j)$.

So I worked inside out, and split the inside sum into two as such:

$2\sum_{j=1}^{10} i + 2\sum_{j=1}^{10} j$

Second one I used the formula $n(n+1)/2$
and got $55 *2 = 110$

First One is just $2i$

So then I placed that into the outside sum and got:

$\sum_{i=1}^{8} {(110+2i)}$

$2 \sum_{i=1}^{8} i + 110 \sum_{i=1}^{8}$

Which is $n(n+1)/2$

$36*2 = 72 + 110 = 182$

Is 182 the correct answer here? I'm not sure if I'm doing this correctly.

Answer 1

You haven´t considered $\sum_{j=1}^{10} 2i=20i$.

In total it is $\sum_{i=1}^{8} (20i+110)=20\sum_{i=1}^{8} i+110\sum_{i=1}^{8} 1$

$=\frac{8\cdot 9}{2}\cdot 20+110\cdot 8= 720+880=1600$

Edit: Pay attention on $\sum_{i=1}^8 110=110\cdot \sum_{i=1}^8 1$ and $\sum_{i=1}^8 1=8$, it is not 1.

Answer 2

I am afraid that is not right. Another way to do it may be $$ \sum\limits_{i=1}^{8}\sum\limits_{j=1}^{10}(2i + 2j)=2\sum\limits_{i=1}^{8}\sum\limits_{j=1}^{10}i + 2\sum\limits_{i=1}^{8}\sum\limits_{j=1}^{10}j=20\sum\limits_{i=1}^{8}i + 16\sum\limits_{j=1}^{10}j=\\=20\frac{8*9}{2} + 16\frac{10*11}{2}=720+880=1600 $$