Given $f(n)=\sum_{k=0}^{n}2^k$, use mathematical induction to prove $f(n)=2^{n+1}-1$

Question

For $n\in\mathbb{N}$ the funktion is defined as $f(n)=\sum_{k=0}^{n}2^k$

I have to use induction to show that $f(n)=2^{n+1}-1$

I have made the first step and shown that $f(n_0)$ is true for $n_0\in\mathbb{N}$ since both sides equals 1.

But I can't figure out how to show that it is true for $f(n+1)$.

score -1 · Answer 1 · answered Nov 15 '16 at 12:02

-1

Assume $f(n) = 2^{n+1} - 1$ is true for some $n \in \mathbb{N}$. If $f(n+1) = 2^{n+2} - 1$ holds, then we are done. Notice

$$ \sum_{k=0}^{n+1} 2^k = f(n) + 2^{n+1} = 2^{n+1} -1 + 2^{n+1} = 2^{n+2} -1 $$

answered Nov 15 '16 at 12:02

ILoveMath

why the downvote? – ILoveMath Nov 15 '16 at 12:17
1

I didn't downvote, but people may think that this question did not deserve an answer. It is a duplicate and the requester did not show any effort. – Pieter21 Nov 15 '16 at 12:39
Upvoted because in my honest opinion a correct and helpful answer doesn't deserve a negative score – Szmagpie Nov 25 '16 at 13:24

1 Answers1