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Calculate the sum:

$$ \sum_{k=0}^n (-1)^k {n+1\choose k+1} $$

I don't know if I'm so tired or what, but I can't calculate this sum. The result is supposed to be $1$ but I always get something else...

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  • 5
    $\begingroup$ Hint: Use the binomial theorem to expand $(1-1)^{n+1}$. $\endgroup$ – Henning Makholm Jul 29 '12 at 21:42
  • $\begingroup$ $ (x + 1)^n = \sum_{k = 0}^{n} {n \choose k} x^k. $ $\endgroup$ – user2468 Jul 29 '12 at 21:47
  • $\begingroup$ Oh, @Henning: I typed my answer up, but it turns out I'm 5 minutes after your (earlier posted, yet containing more or less everything in my answer) comment. $\endgroup$ – davidlowryduda Jul 29 '12 at 21:50
  • $\begingroup$ possible duplicate of Evaluate $ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots$ $\endgroup$ – Grigory M Jan 9 '14 at 9:06
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Using the binomial theorem we have: $$ (1 + (-1))^{n+1} = {{n+1} \choose 0} (-1)^0 + {{n+1} \choose 1} (-1)^1 + \ldots + {{n+1} \choose {n+1}} (-1)^{n+1}.$$ "Divide" by ${-1}$ to get: $$ - (1 - 1)^{n+1} = -{{n+1} \choose 0} + \color{blue}{{{n+1} \choose 1} (-1)^0 + \ldots + {{n+1} \choose {n+1}} (-1)^{n}}.$$ This pretty much solves it.

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HINT

  1. $\displaystyle (1-1)^{n+1} = \sum_{k = 0}^{n+1} {n \choose k}1^k(-1)^{n-k}$
  2. What terms are or aren't in your sum that are in the one above?
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2
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Another way to see it: prove that $$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\,\,\,,\,\,\text{so}$$ $$\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}=\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k}+\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k+1}=$$ $$=(1+(-1))^n-\sum_{k=0}^n(-1)^{k+1}\cdot 1^{n-k-1}\binom{n}{k+1}=0-(1-1)^n+1=1$$

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1
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$$\sum_{k=0}^n (-1)^k {n+1\choose k+1}=-\sum_{k=0}^n (-1)^{k+1} {n+1\choose k+1}= $$ $$=-\left(\sum_{k=0}^n (-1)^{k+1} {n+1\choose k+1}\right)=$$ $$=-\left(\sum_{j=1}^{n+1} (-1)^{j} {n+1\choose j}+(-1)^{0} {n+1\choose 0}-(-1)^{0} {n+1\choose 0}\right)= $$ $$=-\left(\sum_{j=0}^{n+1} (-1)^{j} {n+1\choose j}-(-1)^{0} {n+1\choose 0}\right)=-(1-1)^{n+1}+1=1$$

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