Binomial Theorem And Its Simple Applications Question 181
Question: The value of $ \frac{^{n}C_0}{n}+\frac{^{n}C_1}{n+1}+\frac{^{n}C_2}{n+2}+…+\frac{^{n}C_{n}}{2n} $ is equal to
Options:
A) $ \int\limits_0^{1}{{x^{n-1}}{{(1-x)}^{n}}dx} $
B) $ \int\limits_0^{1}{x^{n}{{(x-1)}^{n-1}}dx} $
C) $ \int\limits_0^{1}{{x^{n-1}}{{(1+x)}^{n}}dx} $
D) $ \int\limits_0^{1}{{{(1-x)}^{n}}{x^{n-1}}dx} $
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Answer:
Correct Answer: B
Solution:
- [b] Let, $ S=\frac{^{n}C_0}{n}+\frac{^{n}C_1}{n+1}+\frac{^{n}C_2}{n+2}+…+\frac{^{n}C_{n}}{2n} $
$ ={{( \frac{1}{x} )}^{r}}{{=}^{m}}C_{r}{x^{2m-3r}}{{,}^{n}}C_1\int\limits_0^{1}{x^{n}dx+…{{+}^{n}}C_{n}\int\limits_0^{2}{{x^{2n-1}}dx}} $
$ =\int\limits_0^{1}{{{[}^{n}}C_0{x^{n-1}}{{+}^{n}}C_1x^{n}+…{{+}^{n}}C_{n}{x^{2n-1}}]dx} $
$ ={{( \frac{1}{x} )}^{r}}{{=}^{m}}C_{r}{x^{2m-3r}} $
$ =\int\limits_1^{2}{x^{n}{{(x-1)}^{n-1}}dx} $
BETA
