Proof by induction that $\sum_{j=0}^n 2^j = 2^{n+1} - 1$

Question

I am trying to solve a previous test for an exam, and there are no solutions. The problem I am trying to solve is

If $n$ is a natural number, then $1 + 2 + 2^2 + 2+3 + ... + 2^n = 2^{n+1} -1$

$$\forall n \in\mathbb N,\,\sum\limits_{j=0}^n{2^j} = 2^{n+1} - 1$$

I am pretty sure this is a proof by induction, but I can only get so far.

Let $P(n)$ be $\sum\limits_{j=0}^n{2^j} = 2^{n+1} - 1$

Prove $P(1)$:
$3 = 3$, True.

Prove $\forall n \in\mathbb N,\,P(n) \implies P(n+1)$: Assume $n \in \mathbb N$
Then, $n = 0 \lor n > 0$

Case $1$: Assume $n = 0$:
Then $1 = 1$, True.

Case $2$: Assume $n > 0$:
... I am stuck actually proving that $P(n) \implies P(n + l)$

See also: http://math.stackexchange.com/questions/22599/how-do-i-prove-this-by-induction-sum-of-powers-of-2 — Martin Sleziak, Feb 19 '15 at 09:34

score 1 · Answer 1 · answered Apr 20 '14 at 01:02

1

Let $S_n=\sum_{j=0}^n 2^j$. Then,

$$S_{n+1}=S_n+2^{n+1}=2^{n+1}-1+2^{n+1}=2^{n+2}-1$$

answered Apr 20 '14 at 01:02

ajotatxe

Thanks thats very helpful! I am not totally sure how $S_{n+1} = S_n + 2^{n+1}$ I may have to review sum laws. – user139934 Apr 20 '14 at 01:27
Nvm i figured it out thanks for the help! – user139934 Apr 20 '14 at 02:09

score 1 · Accepted Answer · answered Apr 20 '14 at 01:35

1

Suppose the case $P(n)$ holds. That is, we have

$$ 1 + 2^1 + 2^2 + ... + 2^n = 2^{n+1} - 1 $$

We want to show the case $P(n+1)$ is implied by $P(n)$. Notice

$$ 1 + 2^1 + 2^2 + ... + 2^n + 2^{n+1} = 2^{n+1} - 1 + 2^{n+1} = 2 \cdot 2^{n+1} - 1 = 2^{n+2} - 1$$

Hence $P(n) \implies P(n+1)$, and the problem is now solved by the principle of mathematical induction.

Hope this helps.

answered Apr 20 '14 at 01:35

Thanks for your response, I must be missing something, I just dont know P(n) = 1 + 2 +... $2^n + 2^{n+1}$, i thought it was just up to n, not n+1. – user139934 Apr 20 '14 at 01:51
Nvm i figured it out thanks for the help! – user139934 Apr 20 '14 at 02:09

2 Answers2