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Consider the coefficients of the Binomial Expansion, $(1-x)^n$. Now, if I say, for some $k$, where $k$ is odd, I truncate the sum. Prove that when $k$ is odd this sum is non-positive, and that when $k$ is even the sum is non-negative. More precisely, I have $$(1-x)^n = \binom{n}{0} -\binom{n}{1}x + \cdots +(-1)^k\binom{n}{k}x^k +\cdots + (-1)^{n}\binom{n}{n}x^n$$ Now consider the following sum:

$$ \binom{n}{0} -\binom{n}{1} + \cdots + (-1)^k\binom{n}{k}$$ Prove that this is nonnegative if $k$ is even, and nonpositive if $k$ is odd.

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    $\begingroup$ You should be able to find a closed formula for this sum. $\endgroup$ Mar 8 '18 at 18:08
  • $\begingroup$ Intuitively that was what I was trying to do, and I never tried doing Jaideep Khare did. I was stuck in taking binomial coeff. common and was stuck. Anyways, doing that way did give me a series, that was neither an AP, GP or an AGP - I was essentially stuck here. Thanks a lot for the suggestions $\endgroup$ Mar 8 '18 at 19:17
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You have

$${n \choose k}={n-1 \choose k-1}+{n-1\choose k}$$ Now you're computing an alternating sum, so you get a telescoping effect:

$${n\choose 0}-{n\choose 1}+\cdots+(-1)^k{n\choose k}=$$ $$=\bigg(0+{n-1\choose 0}\bigg)-\bigg({n-1\choose 0}+{n-1\choose 1}\bigg)+\cdots + (-1)^k\bigg({n-1 \choose k-1}+{n-1\choose k}\bigg)$$ $$=\left[{n-1\choose 0}-{n-1\choose 0}\right ]+\left[{n-1\choose 1}-{n-1\choose 1}\right]+\cdots + \left[{n-1 \choose k-1}-{n-1\choose k-1}\right]+(-1)^k{n-1\choose k}=$$ $$=\color{blue}{(-1)^k {n-1\choose k}}$$

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