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Prove for all integers $n \geq 1$

${n\choose 0} - \frac12 {n \choose 1} + \frac{1}{2^2} {n \choose 2} - \frac{1}{2^3} {n\choose 3} + \cdots + (-1)^{n-1}\frac{1}{2^{n-1}} {n \choose n-1} = \left\{ \begin{array}{ll} 0 & \text{if n is even} \\ \frac{1}{2^{n-1}} & \text{if n is odd} \end{array} \right\} $

I've tried Binomial Theorem and Pascal's formula. Any help would be appreciated.

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  • $\begingroup$ I have edited your post. Please check that the edit is correct. $\endgroup$ – Cameron Williams Apr 1 '14 at 15:45
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Substitute $x=\frac{-1}{2}$ in the binomial expansion for $(1+x)^n$

Hint: Expand the formula, your sum is everything but the last term.

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HINT:

$$\binom n0+\binom n1\left(-\frac12\right)^1++\binom n2\left(-\frac12\right)^2+\cdots+\binom nr\left(-\frac12\right)^r+\cdots++\binom n{n-1}\left(-\frac12\right)^{n-1}+\left(-\frac12\right)^n$$

$$=\left[1+\left(-\frac12\right)\right]^n$$

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