Binomial Theorem And Its Simple Applications Question 147
Question: The value of $ \frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+….. $ is equal to
[Karnataka CET 2000]
Options:
A) $ \frac{2^{n}-1}{n+1} $
B) $ n{{.2}^{n}} $
C) $ \frac{2^{n}}{n} $
D) $ \frac{2^{n}+1}{n+1} $
Show Answer
Answer:
Correct Answer: A
Solution:
- We know that $ \frac{{{(1+x)}^{n}}-{{(1-x)}^{n}}}{2}=C_1x+C_3x^{3}+C_5x^{5}+…. $
Integrating from $ x=0 $ to $ x=1 $ , we get $ \frac{1}{2}\int\limits_0^{1}{{{{(1+x)}^{n}}-{{(1-x)}^{n}}},}dx $
$ =\int\limits_0^{1}{(C_1x+C_3x^{3}+C_5x^{5}+….)}dx $
therefore $ \frac{1}{2}{ \frac{{{(1+x)}^{n+1}}}{n+1}+\frac{{{(1-x)}^{n+1}}}{n+1} }_0^{1}=\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+…. $ or $ \frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+….=\frac{1}{2}{ \frac{{2^{n+1}}-1}{n+1}+\frac{0-1}{n+1} } $
$ =\frac{1}{2}( \frac{{2^{n+1}}-2}{n+1} )=\frac{2^{n}-1}{n+1} $
BETA
