TITLE: Binomial Theorem Question..
QUESTION [5 upvotes]: 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!

REPLY [3 votes]: $$(1+1)^{41}= \sum^{41}_{k=0} \binom{41}{k}$$
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 
$$2^{40}= \sum^{20}_{k=0} \binom{41}{k}$$