calcuate $\sum_{i=0}^{n} 2^{2i}$

Question

I want to calcuate this problem: $\sum_{i=0}^{n} 2^{2i+5}$

I know that we can expand this problem like this:

$\sum_{i=0}^{n} (2^{2i+5})$

$=\sum_{i=0}^{n} (2^5 \times 2^{2i})$

$=\sum_{i=0}^{n} (32 \times 2^{2i})$

$=32\sum_{i=0}^{n} (2^{2i})$

But I stopped here. is there any way to calcuate $\sum_{i=0}^{n} (2^{2i})$ ?

Answer 1

Thanks for Slade, I realized $2^{2\times i}=(2^{2})^{i}=4^{i}$

So,

$=32\sum_{i=0}^{n} (2^{2i}) = 32\sum_{i=0}^{n} 4^{i} = 32 \times \frac{(4^{n-1})-1}{4-1}=32 \times \frac{(4^{n-1})-1}{3}$

(Using Closed-form formula for the sum of a geometric series.)