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!