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anonymous

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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    it should be mulitiplied...

  2. anonymous
    • 4 years ago
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    http://letmegooglethat.com/?q=+Evaluwate+n%2Bk+choose+2k

  3. anonymous
    • 4 years ago
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    did that help

  4. anonymous
    • 4 years ago
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    I'm going to check it and then let you know... thanks

  5. JamesJ
    • 4 years ago
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    You want \[ \sum_{k=0}^n (-4)^k { n+k \choose 2k } \] yes?

  6. anonymous
    • 4 years ago
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    yes

  7. JamesJ
    • 4 years ago
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    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
    • 4 years ago
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    If you calculate this further, you'll find n = 3, sum = -7 n = 4, sum = 9 n = 5, sum = -11

  9. JamesJ
    • 4 years ago
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    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
    • 4 years ago
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    Thanks, we were just confused with what the sum would be equal to...I'll get to work on that

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