Binomial Theorem And Its Simple Applications Question 148
Question: The sum to $ (n+1) $ terms of the following series $ \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+ $ ….. is
Options:
A) $ \frac{1}{n+1} $
B) $ \frac{1}{n+2} $
C) $ \frac{1}{n(n+1)} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
- $ {{(1-x)}^{n}}=C_0-C_1x+C_2x^{2}-C_3x^{3}+….. $
therefore $ x{{(1-x)}^{n}}=C_0x-C_1x^{2}+C_2x^{3}-C_3x^{4}+….. $
therefore $ \int\limits_0^{1}{x}{{(1-x)}^{n}}dx=\int\limits_0^{1}{(C_0x-C_1x^{2}+C_2x^{3}….)dx} $ ……………(i) The integral on the LHS $ =\int\limits_1^{0}{(1-t)t^{n}(-dt),} $ by putting $ 1-x=t $
$ =\int\limits_0^{1}{(t^{n}-{t^{n+1}})},dt=\frac{1}{n+1}-\frac{1}{n+2} $ Whereas the integral on the RHS of (i) $ =[ \frac{C_0x^{2}}{2}-\frac{C_1x^{3}}{3}+\frac{C_2x^{4}}{4}-…. ] $
$ =\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-…. $
$ \therefore \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-…. $ to $ (n+1) $ terms $ =\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)} $ .
BETA
