Binomial Theorem And Its Simple Applications Question 178
Question: If $ C_0,C_1,C_2,…….,C_{n} $ are the binomial coefficients, then $ 2.C_1+2^{3}.C_3+2^{5}.C_5+…. $ equals
[AMU 1999]
Options:
A) $ \frac{3^{n}+{{(-1)}^{n}}}{2} $
B) $ \frac{3^{n}-{{(-1)}^{n}}}{2} $
C) $ \frac{3^{n}+1}{2} $
D) $ \frac{3^{n}-1}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+C_3x^{3}+…..+C_{n}x^{n} $
$ {{(1-x)}^{n}}=C_0-C_1x+C_2x^{2}-C_3x^{3}+…..+{{(-1)}^{n}}C_{n}x^{n} $
$ [{{(1+x)}^{n}}-{{(1-x)}^{n}}]=2,[C_1x+C_3x^{3}+C_5x^{5}+…] $
$ \frac{1}{2}[{{(1+x)}^{n}}-{{(1-x)}^{n}}]=C_1x+C_3x^{3}+C_5x^{5}+……. $ Put x = 2, $ 2.C_1+2^{3}.C_3+2^{5}.C_5+…..,=\frac{3^{n}-{{(-1)}^{n}}}{2} $
BETA
