anonymous
  • anonymous
Evaluate \[\sum_{k=0}^{n}(-4)^{k}\](n+k choose 2k) in closed form.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
it should be mulitiplied...
anonymous
  • anonymous
http://letmegooglethat.com/?q=+Evaluwate+n%2Bk+choose+2k
anonymous
  • anonymous
did that help

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anonymous
  • anonymous
I'm going to check it and then let you know... thanks
JamesJ
  • JamesJ
You want \[ \sum_{k=0}^n (-4)^k { n+k \choose 2k } \] yes?
anonymous
  • anonymous
yes
JamesJ
  • 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: \[ \Sigma = (-4)^0 { 2 \choose 0} + (-4)^1 { 3 \choose 2} + (-4)^2 { 4 \choose 4} \] \[ = 1 -4(3) + 16(1) = 5 \]
JamesJ
  • JamesJ
If you calculate this further, you'll find n = 3, sum = -7 n = 4, sum = 9 n = 5, sum = -11
JamesJ
  • 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.
anonymous
  • anonymous
Thanks, we were just confused with what the sum would be equal to...I'll get to work on that

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