So I have a homework assignment that has brought me great strain over the past 2 days. No video or online example have been able to help me with this issue either and I don't know where to turn.

I’m given



After writing the first 6 terms of the series: 0, 1, 3, 7, 15, 31, 63 I come up with an alternate formula of


I then have to prove these formulas are the same using Induction in 3 parts:

  • Proving the base case
  • Stating my Inductive Hypothesis
  • Showing the Inductive Step

I have done Inductive proofs before but I don’t know how to show cases or do manipulations on a recursive formula. I don’t know how to represent when n = k then n = k + 1 or showing the approach by using n = k – 1 then n = k.

Any ideas?


2 Answers 2


For the setup, we need to assume that $a_n = 2^n - 1$ for some $n$, and then show that the formula holds for $n + 1$ instead. That is, we need to show that $$a_{n + 1} = 2^{n + 1} - 1$$

Let's just compute directly:

\begin{align*} a_{n + 1} &= 2a_n + 1 \hspace{1.55in}\text{// recursion relation} \\ &= 2 \cdot (2^n - 1) + 1 \hspace{1in} \text{// induction hypothesis} \\ &= 2^{n + 1} - 2 + 1 \hspace{1.15in} \text{// arithmetic} \\ &= 2^{n + 1} - 1 \end{align*}

which is exactly what we wanted to be true.

  • 2
    $\begingroup$ +1, great answer. Just as an aside to this fine answer, I'm not sure what your background is @Chris, but what you've observed here falls under the more general problem of solving linear recurrence relations. Here is a nice resource which describes how you can generally solve recurrence relations of this kind: eecs.yorku.ca/course_archive/2008-09/S/1019/Website_files/…. You can find many more like this by simply searching "solving linear recurrence relations." $\endgroup$ Oct 23, 2013 at 1:37
  • $\begingroup$ You had an oops with your arithmetic when you distributed the 2. But anyway, thank you very much, I can't believe it was so simple or that you could represent things that way. The challenge comes when trying to figure out what side to prove and which side I can manipulate to look like the other side. $\endgroup$
    – Chris
    Oct 23, 2013 at 1:48

Proving the base case should be rather simple.

For the inductive hypothesis, we'll assume that for $k\geq1$, $$a_{k-1}=2^{k-1}-1$$ From this you need to prove that $a_k=2^k-1$. It shouldn't be too tough to get it from here just by following the recurrence relation.


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