Evaluate -122011-k-01006-1k 3k 2012C2kcorrect answer - 5- wrong answer 0

Consider the complex number ω=cosπ3+isinπ3. Using the binomial theorem, nbsp; nbsp; Re(ω^2012)=Re(cosπ3+isinπ3)^2012 =Re (12+i√(3)2)^2012 =(12)^2012 ^2012C_ ...

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Standard XIII

Mathematics

De-Moivre's Theorem

Evaluate -122...

Question

Evaluate $\frac{- 1}{2^{2011}} 1006 \sum k = 0 (- 1)^{k} 3^{k}^{2012} C_{2 k}$
(correct answer + 5, wrong answer 0)

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Solution

Consider the complex number $ω = cos \frac{π}{3} + i sin \frac{π}{3}$ .
Using the binomial theorem,
$R e (ω^{2012}) = R e {(cos \frac{π}{3} + i sin \frac{π}{3})}^{2012}$
$= R e {(\frac{1}{2} + i \frac{\sqrt{3}}{2})}^{2012}$
$= {(\frac{1}{2})}^{2012} -^{2012} C_{2} {(\frac{1}{2})}^{2010} (\frac{3}{2^{2}}) +^{2012} C_{4} {(\frac{1}{2})}^{2008} (\frac{3^{2}}{2^{4}}) + \dots + (\frac{3^{1006}}{2^{2012}})$
$= \frac{1}{2^{2012}} [1 - 3 (^{2012} C_{2}) + 3^{2} (^{2012} C_{4}) + \dots + 3^{1006} (^{2012} C_{2012})]$

On the other hand, using De Moivre's theorem,
$R e (ω^{2012}) = R e (cos \frac{2012 π}{3} + i sin \frac{2012 π}{3}) = cos \frac{2012 π}{3} = \frac{- 1}{2}$
Thus, $\frac{1}{2^{2012}} [1 - 3 (^{2012} C_{2}) + 3^{2} (^{2012} C_{4}) - \dots + 3^{1006} (^{2012} C_{2012})] = \frac{- 1}{2}$
$∴ \frac{- 1}{2^{2011}} [1 - 3 (^{2012} C_{2}) + 3^{2} (^{2012} C_{4}) - \dots + 3^{1006} (^{2012} C_{2012})] = 1$

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Evaluate −1220111006∑k=0(−1)k 3k 2012C2k (correct answer + 5, wrong answer 0)

Evaluate $\frac{- 1}{2^{2011}} 1006 \sum k = 0 (- 1)^{k} 3^{k}^{2012} C_{2 k}$
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