Question

If $C_{0}$, $C_{1}$, $C_{2}$, $C_{3}$, $\cdots$ are binomial coefficients in the expansion of $(1 + x)^n$, then $\frac{C_{0}}{3}- \frac{C_{1}}{4}+ \frac{C_{2}}{5}- \frac{C_{3}}{6}+ \cdots$ is equal to

• A. $\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}$
• B. $\frac{1}{n+1}+\frac{2}{n+2}-\frac{1}{n+3}$
• C. $\frac{1}{n+2}-\frac{1}{n+2}+\frac{1}{n+3}$
• D. $\frac{2}{n+1}-\frac{1}{n+2}+\frac{2}{n+3}$

Answer 1

The Correct Option is A

We know
$(1-x)^{n}=C_{0}-C_{1} x+C_{2} x^{2}-\ldots+(-1)^{n}\left(C_{n} \cdot x^{n}\right.$
On multiplying both sides by $x^{2}$, we get
$(1-x)^{n} x^{2}=C_{0} x^{2}-C_{1} x^{3}+C_{2} x^{4}+\ldots$
On integrating both sides by taking limit 0 to 1
$\therefore \int_{0}^{1}(1-x)^{n} x^{2} d x $
$=\int_{0}^{1}\left(C_{0} x^{2}-C_{1} x^{3}+C_{2} x^{4}+\ldots\right) d x $
$\int_{0}^{1} x^{n}(1-x)^{2} d x $
$=\left[C_{0} \frac{x^{3}}{3}-C_{1} \frac{x^{4}}{4}+C_{2} \frac{x^{5}}{5}-\ldots\right]_{0}^{1}$
$\Rightarrow \int_{0}^{1} x^{n}\left(1+x^{2}-2 x\right) d x=\frac{C_{0}}{3}-\frac{C_{1}}{4}+\frac{C_{2}}{5}-\ldots$
$\Rightarrow \frac{C_{0}}{3}-\frac{C_{1}}{4}+\frac{C_{2}}{5}-\ldots$
$=\int_{0}^{1}\left(x^{n}+x^{n+2}-2 x^{n+1}\right) d x$
$=\left[\frac{x^{n+1}}{n+1}+\frac{x^{n+3}}{n+3}-\frac{2 x^{n+2}}{n+2}\right]_{0}^{1}$
$=\left[\frac{1}{n+1}+\frac{1}{n+3}-\frac{2}{n+2}\right]$