Question

Question #2f54d

Answer 1

Proof by Induction

Base Case (n=1)

`sum_{n=1}^1 4^i=4=4/3(4^1-1)`

Induction Hypothesis (n=k)

Assume: `sum_{i=1}^k4^i=4/3(4^k-1)`

Induction Step (n=k+1)

Show: `sum_{i=1}^{k+1}4^i=4/3(4^{k+1}-1)`

`sum_{i=1}^{k+1}4^i=sum_{i=1}^k4^i+4^{k+1}=4/3(4^k-1)+4/3(3cdot4^k)=4/3(4cdot4^k-1)=4/3(4^{k+1}-1)`

Hence, for all natural number `n`,

`sum_{i=1}^n4^i=4/3(4^n-1)`.

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I hope that this was helpful.