Evaluate $$\sum_{k=0}^{n}(-4)^{k}$$(n+k choose 2k) in closed form.

Question

anonymous · Answer

it should be mulitiplied...

anonymous · Answer

http://letmegooglethat.com/?q=+Evaluwate+n%2Bk+choose+2k

anonymous · Answer

did that help

anonymous · Answer

I'm going to check it and then let you know... thanks

jamesj · Answer

You want
$$ \sum_{k=0}^n (-4)^k { n+k \choose 2k } $$
yes?

anonymous · Answer

yes

jamesj · Answer

Right.  Let's test a few values and see if we can detect a pattern

n=0:
sum = 1

n=1:
$$ \Sigma = (-4)^0 { 0 \choose 0} + (-4)^1 { 2 \choose 2} = 1 - 4 = -3 $$

n=2: 
$$ \Sigma = (-4)^0 { 2 \choose 0}  + (-4)^1 { 3 \choose 2} + (-4)^2 { 4 \choose 4} $$
$$ = 1 -4(3) + 16(1) = 5 $$

jamesj · Answer

If you calculate this further, you'll find
n = 3, sum = -7
n = 4, sum = 9
n = 5, sum = -11

jamesj · Answer

Hence it looks like the sum is equal to
$$ (-1)^n(2n+1) $$

To prove that, now, use induction and/or other identities you have for binomial coefficients.

anonymous · Answer

Thanks, we were just confused with what the sum would be equal to...I'll get to work on that