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

1. anonymous

it should be mulitiplied...

2. anonymous
3. anonymous

did that help

4. anonymous

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

5. JamesJ

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

6. anonymous

yes

7. JamesJ

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: $\Sigma = (-4)^0 { 2 \choose 0} + (-4)^1 { 3 \choose 2} + (-4)^2 { 4 \choose 4}$ $= 1 -4(3) + 16(1) = 5$

8. JamesJ

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

9. JamesJ

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.

10. anonymous

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