Compute \[\frac{-1}{2^{2011}}\sum_{k=0}^{1006}(-1)^k3^k\binom{2012}{2k}\].

This question came from the Singapore Maths Olympiad 2012 Open Section Round 1.

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TopNewestSince, we need only the even binomials , We consider : \(\frac{(1 + x )^{2012} + (1 - x)^{2012}}{2} = \displaystyle \sum_{k=0}^{1006} x^{2k}{2012 \choose 2k} \), Replacing x by ix in the above identity, \(\frac{(1 + ix )^{2012} + (1 - ix)^{2012}}{2} = \displaystyle \sum_{k=0}^{1006} (-1)^kx^{2k}{2012 \choose 2k} \) Replacing x = \( \sqrt{3} , \\ \frac{(1 + i\sqrt{3} )^{2012} + (1 - i\sqrt{3})^{2012}}{2} = \displaystyle \sum_{k=0}^{1006} (-1)^k3^k{2012 \choose 2k} \\ \Rightarrow S = \displaystyle \sum_{k=0}^{1006} (-1)^k3^k{2012 \choose 2k} = \frac{[2e^{i\frac{\pi}{3}}]^{2012} + [2e^{i\frac{-\pi}{3}}]^{2012}}{2} \\ = \frac{2(2^{2012}cos(\frac{2012\pi}{3})}{2} = -2^{2011} \) \( \Rightarrow \)Answer = \( \frac{S}{-2^{2011}} = \frac{-2^{2011}}{-2^{2011}} =\fbox{1} \)

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Nice!

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dear friend, I suppose if I have not done any mistake the answer will be 1. it is pretty much obvious actually. OBSERVE CAREFULLY the expression with the summation sine can be written as,, [[\((1+i\(\sqrt{3}\))^2012)]+[\((1-i\(\(sqrt{3}\))^2012)]]/2 thus the given expression equals, = -[[\(((\frac{1}{2}\)+i\(\frac{\(\sqrt{3}\)}{2}))^2012)]+[\(((\frac{1}{2}\)-i\(\frac{\(\sqrt{3}\)}{2}))^2012)]] -[\(\cos\x\)+i\(\sin\x\) +\(\cos\x\)-i\(\sin\x\)] [where x=\(\frac{2012pi}{3}\)] [applying de moivers theorem] = -2[\(\cos\120\)]=1(answer)

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