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# Binomial Dilemna

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.

Note by Ho Wei Haw
3 years, 11 months ago

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Since, 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 = $$\frac{S}{-2^{2011}} = \frac{-2^{2011}}{-2^{2011}} =\fbox{1}$$ · 3 years, 11 months ago

Nice! · 3 years, 11 months ago

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) · 3 years, 11 months ago