combinatorics - Induction proof of $sum_{k=1}^{n} binom n k = 2^n -1$

Prove by induction: $$\sum_{k=1}^{n} \binom n k = 2^n -1$$ for all $n\in \mathbb{N}$.

Today I wrote calculus exam, I had this problem given.

I have the feeling that I will get $0$ points for my solution, because I did this:

Base Case: $n=1$
$$\sum_{k=1}^{1} \binom 1 1 = 1 = 2^1 -1 .$$

Induction Hypothesis: for all $n \in \mathbb{N}$:
$$\sum_{k=1}^{n} \binom n k = 2^n -1$$

Induction Step: $n \rightarrow n+1$
$$\sum_{k=1}^{n+1} \binom {n+1} {k} = \sum_{k=1}^{n} \binom {n+1} {k} + \binom{n+1}{n+1} = 2^{n+1} -1$$

Please show me my mistake because next time is my last chance in this class.

Answer

Your first mistake is that you stated the induction hypothesis incorrectly: it should be simply that

$$\sum_{k=1}^n\binom{n}k=2^n-1\;.\tag{1}$$

What you gave as induction hypothesis is actually the theorem that you’re trying to prove!

Now, assuming $(1)$ you want to prove that $$\sum_{k=1}^{n+1}\binom{n+1}k=2^{n+1}-1\;.$$

This is an occasion when it’s not a good idea to split off the last term. Instead, use the Pascal’s triangle identity for binomial coefficients:

$$\begin{align*} \sum_{k=1}^{n+1}\binom{n+1}k&=\sum_{k=1}^{n+1}\left(\binom{n}k+\binom{n}{k-1}\right)\\\\ &=\sum_{k=1}^{n+1}\binom{n}k+\sum_{k=1}^{n+1}\binom{n}{k-1}\\\\ &\overset{*}=\sum_{k=1}^n\binom{n}k+\sum_{k=0}^n\binom{n}k\\\\ &=\left(2^n-1\right)+\binom{n}0+\sum_{k=1}^n\binom{n}k\\\\ &=2^n-1+1+2^n-1\\\\ &=2\cdot2^n-1\\\\ &=2^{n+1}-1\;. \end{align*}$$

At the starred step I used the fact that $\binom{n}{n+1}=0$ to drop the last term of the first summation, and I did an index shift, replacing $k-1$ by $k$, in the second summation.