Complex Numbers And Quadratic Equations question 146
Question: If $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+…..+C_{n}x^{n}, $ then the value of $ C_0-C_2+C_4-C_6+….. $ is
Options:
A) $ 2^{n} $
B) $ 2^{n}\cos \frac{n\pi }{2} $
C) $ 2^{n}\sin \frac{n\pi }{2} $
D) $ {2^{n/2}}\cos \frac{n\pi }{4} $
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Answer:
Correct Answer: D
Solution:
Since $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+…..+C_{n}x^{n} $ Put $ x=i $ , on both the sides, we get $ {{(1+i)}^{n}}=(C_0-C_2+C_4-…..)+i(C_1-C_3+C_5-…..) $ …..(i) Also, $ 1+i=\sqrt{2}( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} ) $ in amplitude modulus form
Þ $ {{(1+i)}^{n}}={2^{n/2}}{{( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} )}^{n}} $ $ ={2^{n/2}}( \cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4} ) $ ….(ii) Equating the real parts in (i) and (ii) we get, $ C_0-C_2+C_4-C_6+…..={2^{n/2}}\cos \frac{n\pi }{4} $
BETA
