Binomial Theorem And Its Simple Applications Question 167
Question: The value of $ ^{4n}C_0{{+}^{4n}}C_4{{+}^{4n}}C_8+….{{+}^{4n}}C_{4n} $ is
Options:
A) $ {2^{4n-2}}+{{(-1)}^{n}}{2^{2n-1}} $
B) $ {2^{4n-2}}+{2^{2n-1}} $
C) $ {2^{2n-1}}+{{(-1)}^{n}},{2^{4n-2}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- We have $ ^{4n}C_0+{{,}^{4n}}C_2x^{2}+{{,}^{4n}}C_4x^{4}+…+{{,}^{4n}}C_{4n}x^{4n} $
$ =\frac{1}{2}[{{(1+x)}^{4n}}+{{(1-x)}^{4n}}] $ Putting $ x=1 $ and x = i, we get $ ^{4n}C_0+{{,}^{4n}}C_2+{{,}^{4n}}C_4+…+{{,}^{4n}}C_{4n}=\frac{1}{2}[2^{4n}] $ and $ ^{4n}C_0-{{,}^{4n}}C_2+{{,}^{4n}}C_4-…+{{,}^{4n}}C_{4n} $ = $ \frac{1}{2}[{{(1+i)}^{4n}}+{{(1-i)}^{4n}}] $
Thus, $ 2{{[}^{4n}}C_0+{{,}^{4n}}C_4+…+{{,}^{4n}}C_{4n}] $
$ ={2^{4n-1}}+\frac{1}{2}{{[{{(1+i)}^{4n}}+(1-i)]}^{4n}} $ Now, $ {{(1+i)}^{4n}}+{{(1-i)}^{4n}}={{[ \sqrt{2}( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} ) ]}^{4n}} $
$ +{{[ \sqrt{2}( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} ) ]}^{4n}} $
$ =2^{2n}(\cos n\pi +i\sin n\pi )+2^{2n}(\cos n\pi -i\sin n\pi ) $
$ ={2^{2n+1}}\cos n\pi ={2^{2n+1}}{{(-1)}^{n}} $
$ \therefore $ $ 2{{[}^{4n}}C_0+{{,}^{4n}}C_4+…+{{,}^{4n}}C_{4n}]={2^{4n-1}}+\frac{1}{2}{2^{2n+1}}{{(-1)}^{n}} $
therefore $ ^{4n}C_0+{{,}^{4n}}C_4+…+{{,}^{4n}}C_{4n}={2^{4n-2}}+{{(-1)}^{n}}{2^{2n-1}} $ Trick: Check by putting n = 1, 2.
BETA
