Compute Using Binomial Theorem

Question

$$\sum_{k=1}^{10} \binom{10}{k} $$

I know the answer is $2^{10} - 1$ but I don't know how to get to the answer.

Answer 1

Consider the following:

• $\binom{10}{0}$ is the number of ways to pick 0 objects out of 10 objects. (There's just 1 way to do this).
• $\binom{10}{1}$ is the number of ways to pick 1 object out of 10 objects. (10 ways).
• $\binom{10}{2}$ is the number of ways to pick 2 objects out of 10 objects. (45 ways).
• ...
• $\binom{10}{10}$ is the number of ways to pick 10 objects out of 10 objects. (Again, just 1 way).

So $\sum_{k=0}^{10} \binom{10}{k}$ is the total number of choices of any number of objects from a set of 10 objects. That's $2^{10}$, because for each object you have a binary decision: pick it or not.

Thus $\sum_{k=0}^{10} \binom{10}{k} = 2^{10}$. Subtracting $\binom{10}{0} = 1$ you get the desired result: $\sum_{k=1}^{10} \binom{10}{k} = 2^{10}-1$.

Answer 2

Hint: The binomial theorem gives you a formula for $(x+y)^{n}$. In your case, you know the answer should be $2^{10}$... so what could you pick $x$ and $y$ to be so that $(x+y)^{10}=2^{10}$?

EDIT: I thought the sum started at $0$... in this case, you'll have to subtract the zero term $(1)$.