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I am stuck on proving this statement. I know I can use the definition of the geometric sum to prove this, but I wanted to know how to prove this without external definitions.

Let $P(n)$ be the statement $\forall n \in \mathbb N, \sum_{i=0}^{n-1} 2^i = 2^n - 1$

Proof. Let $n \in \mathbb N$.

Base Case (Let $n = 1$)

$\sum_{i=0}^{1-1} 2^i = 2^1 - 1$

$1 = 1$

Therefore $P(1)$ holds

Induction Step: Let $k \in \mathbb N$

Assume P(k): $\sum_{i=0}^{k-1} 2^ i = 2^k - 1$ is true (Induction Hypothesis)

W.T.P. $P(k + 1): \sum_{i=0}^k 2^i = 2^{k+1} - 1$

Now to prove $P(k+1)$, would I let $2^k$ be equal to $\sum_{i=0}^{k-1} (2^i +1)$, by the induction hypothesis? I'm not sure how to proceed.

Rushabh Mehta
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Shrek.Spongebob
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1 Answers1

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If you want a proof by induction, then it is convenient to define some notation as follows. Let $$f(n) = \sum_{i=0}^{n-1} 2^i.$$ We aim to show $f(n) = 2^n - 1$ for all positive integers $n$. To this end, we observe $$f(n+1) = \sum_{i=0}^n 2^i = \sum_{i=0}^{n-1} 2^i + 2^n = f(n) + 2^n.$$ So, if $P(n)$ is the claim that $f(n) = 2^n - 1$, then $$f(n+1) = f(n) + 2^n \overset{P(n)}{=} (2^n - 1) + 2^n = 2 \cdot 2^n - 1 = 2^{n+1} - 1,$$ which completes the induction step. I have omitted establishing the base case $n = 1$ as this step you have already performed.

heropup
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