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This is a problem from a book with no solution.

Show that: $$\sum_{i=0}^{n-1} 2^i = 2^n - 1$$

without using the formula for geometric series.

My lengthy solution is as follows:

Try with $n = 1, 2, 3$ and prove that it works. Which it does.

$$\sum_{i=0}^{n-1} 2^i = 2^n - 1$$ for $n = {1,2,3}$

Assume that it works for $n = p$.

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

Show that it works for $n = p + 1$

$$\begin{align*} LHS_{p+1} =& \sum_{i=0}^p 2^i\\ =& \sum_{i=0}^{p-1} 2^i + 2^p\\ =& LHS_p + 2^p\\ =& RHS_p + 2^p\\ =& 2^p - 1 + 2^p\\ =& 2(2^p) - 1\\ =& 2^{p+1} - 1\\ =& RHS_{p+1} \end{align*}$$

And they match so they must be the same. Did I do this right? :)

Martin Sleziak
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onthecatwalk
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5 Answers5

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Your induction proof is fine, though you need only one initial case.

Here’s a short solution:

$$2\sum_{k=0}^{n-1}2^k=\sum_{k=0}^{n-1}2^{k+1}=\sum_{k=1}^n2^k=\sum_{k=0}^{n-1}2^k+2^n-1\;,$$

so $$\sum_{k=0}^{n-1}2^k=2^n-1\;.$$

(This is essentially proving the formula for the sum of a finite geometric series, but just in this special case.)

Brian M. Scott
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You are proving it by induction and have the correct approach. You only have to try one base case. Then instead of $LHS_{p+1}$ it would be better to use the sum, so
Assume $\sum_{i=0}^{p-1} 2^i = 2^p - 1$ then

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

and you are done

Ross Millikan
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Your proof works just fine. There's also a combinatorial proof expressing the same principle: $2^{i-1}$ is the number of non-zero $i$-digit binary numbers (since any such begins with '1' and has $i-1$ 'free' digits after), so $\sum_{i=1}^{n} 2^{i-1}$ (or equivalently $\sum_{j=0}^{n-1}2^j$) is the total number of (non-zero) binary numbers of $n$ or fewer binary digits - which there are clearly $2^n-1$ of (since they're all the numbers less than $2^n$).

Steven Stadnicki
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Sure that works. With proofs by induction, you don't need to check so many initial cases. Just checking it for the $n = 1$ case is more than enough.

Cheyne H
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Hint: You can avoid induction entirely if you observe that $$\sum_{i=0}^{n-1}2^i=(2-1)\sum_{i=0}^{n-1}2^i$$ and distribute.

Cameron Buie
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