Binomial Theorem Question.. Just studying for my combinatorics exam.  My prof said there would be a question similar to this one on the exam, so I'm trying to sort this one out.
$$\sum^{20}_{k=0} \binom{41}{k}$$
I know if I can factor out $\dbinom{20}{k}$ then I can get the following:
$$A\sum^{20}_{k=0}\binom{20}{k} * 1^k * 1^{n-k} = A(1+1)^{20}$$
I just don't know how to get there...
Any help is greatly appreciated!
 A: $$(1+1)^{41}= \sum^{41}_{k=0} \binom{41}{k}$$
but since the binomial coefficients are symmetrical around $\,k=21\,$ , we know that 
$$2^{41}= \sum^{20}_{k=0}\binom{41}{k}+\sum^{20}_{k=0}\binom{41}{k}\, $$ 
thus 
$$2^{40}= \sum^{20}_{k=0} \binom{41}{k}$$
A: Substitute $x=1$ in the expansion $(1+x)^{41}=\sum_0^{41}\binom{41}{k}x^k$ to obtain $\sum_0^{41}\binom{41}{k}=2^{41}$. Since $\binom{41}{k}=\binom{41}{41-k}$, we get $\sum_0^{20}\binom{41}{k}=\frac{1}{2}\sum_0^{41}\binom{41}{k}=2^{40}$.
A: You’re on the wrong track, I’m afraid. The trick is to realize that your sum is exactly half of the full sum $\sum_{k=0}^{41}\binom{41}k$ because of the symmetry of the binomial coeffients: $\binom{41}{41-k}=\binom{41}k$.
$$\begin{align*}
2^{41}&=\sum_{k=0}^{41}\binom{41}k\\
&=\sum_{k=0}^{20}\binom{41}k+\sum_{k=21}^{41}\binom{41}k\\
&=\sum_{k=0}^{20}\binom{41}k+\sum_{k=21}^{41}\binom{41}{41-k}\\
&=\sum_{k=0}^{20}\binom{41}k+\sum_{i=0}^{20}\binom{41}i&&\text{set }i=41-k\\
&=2\sum_{k=0}^{20}\binom{41}k\;,
\end{align*}$$
so that $$\sum_{k=0}^{20}\binom{41}k=2^{40}\;.$$
