Prove that $\sum_{i=0}^n4^i$ = $1/3(4^{n+1} - 1)$

Question

$\sum_{i=0}^n4^i$ = $1/3(4^{n+1} - 1)$

Attempt:

Let $n =0$ $4^0 = 1 \text{ and } (4^{0+1} -1)/3 = 1$

Assume true at $n = k \text{ so we have} \sum_{i=0}^k4^i = 1/3(4^{k+1} -1)$

The part I'm stuck at is the 3rd step. Can someone point me in the right direction after this?

$\sum_{i=0}^{k+1}4^i$ ?

Here's what I've set the sum to: $$= 1/3(4^{k+1} -1) + 4^{k+1} = 4 * 1/3(4^{k+1})-1 = 1/3(4^{(k+1)+1} - 1$$ Is this correct? — Slae, Oct 27 '14 at 02:17

score 1 · Answer 1 · answered Oct 27 '14 at 02:05

1

Hint: $\displaystyle \sum_{i=0}^{k+1} 4^i = \displaystyle \sum_{i=0}^k 4^i + 4^{k+1}$

answered Oct 27 '14 at 02:05

DeepSea

Do I then set it = to $1/3(4^{(k+1)+1}-1)$ ? – Slae Oct 27 '14 at 02:09
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Yeah, set the sum to $\frac{1}{3}(4^{k+1}-1)$ and add it to $4^{k+1}$ and see what you get... – J. David Taylor Oct 27 '14 at 02:10
@j.DavidTaylor Does what I have look correct? – Slae Oct 27 '14 at 02:23

1 Answers1