I am learning how to evaluate summation and got stuck on evaluating this summation: It would be great if you could help me through this.
$$ \sum_{i=0}^{n} \sum_{i=0}^{m}3^{i+j} $$
also after a bit of scratch work, I have found that this summation will be usefull: $$ 1+r+r^2+r^3+r^4+...+r^n = \frac{1-r^{n+1}}{1-r} $$
Again, any lead into this will help me out a lot! Thanks!
$$ \sum_{i=0}^n \sum_{j=0}^m3^{i+j}$$ $$=\sum_{i=0}^n 3^i \left(\sum_{j=0}^m 3^j\right)\text{ as $i$ is independent of $j$}$$
$$=\sum_{i=0}^n 3^i\left(\frac{3^{m+1}-1}{3-1}\right)$$
$$=\left(\frac{3^{m+1}-1}2\right)\sum_{i=0}^n 3^i\text{ as $m$ is independent of $i,n$}$$
$$=\left(\frac{3^{m+1}-1}2\right)\cdot\left( \frac{3^{n+1}-1}2\right)$$