# Question #5d6db

Oct 22, 2014

Proof by Induction

Base Case (n=1)

${\sum}_{n = 1}^{1} {4}^{i} = 4 = \frac{4}{3} \left({4}^{1} - 1\right)$

Induction Hypothesis (n=k)

Assume: ${\sum}_{i = 1}^{k} {4}^{i} = \frac{4}{3} \left({4}^{k} - 1\right)$

Induction Step (n=k+1)

Show: ${\sum}_{i = 1}^{k + 1} {4}^{i} = \frac{4}{3} \left({4}^{k + 1} - 1\right)$

${\sum}_{i = 1}^{k + 1} {4}^{i} = {\sum}_{i = 1}^{k} {4}^{i} + {4}^{k + 1} = \frac{4}{3} \left({4}^{k} - 1\right) + \frac{4}{3} \left(3 \cdot {4}^{k}\right) = \frac{4}{3} \left(4 \cdot {4}^{k} - 1\right) = \frac{4}{3} \left({4}^{k + 1} - 1\right)$

Hence, for all natural number $n$,

${\sum}_{i = 1}^{n} {4}^{i} = \frac{4}{3} \left({4}^{n} - 1\right)$.

I hope that this was helpful.