Question

$\frac{C_0}{x}-\frac{C_1}{x+1}+\frac{C_2}{x+2}-......+{(-1)}^n\frac{C_n}{x+n}=$_______ where $C_r$ stands for ${{}^{n}C}_r$.

• A. $\frac{n!}{(x+1)...(x+n)}$
• B. $\frac{n!}{x(x+1)...(x+n-1)}$
• C. $\frac{n!}{x(x+1)...(x+n)}$
• D. $\frac{n-1!}{x(x+1)...(x+n)}$

Answer 1

The correct option is C $\frac{n!}{x(x+1)...(x+n)}$

$\frac{c_0}{x}-\frac{c_1}{x+1}+\frac{c_2}{x+2}-............+{(-1)}^n\frac{c_n}{x+n}$

${(1-y)}^n=c_0-c_{1}y+c_2{y}^2-......+c_n{y}^n$

$y^{x-1}{(1-y)}^n=c_0{y}^{x-1}-c_1{y}^x+c_2{y}^{x+1}-......$

${\int }_0^1{y}^{x-1}{(1-y)}^{n}dy={\int }_0^1{c}_0{y}^{x-1}dy-{\int }_0^1{c}_1{y}^{x}dy+..........$

$\Rightarrow {\int }_0^1{y}^{x-1}{(1-y)}^{n}dy=\frac{c_0}{x}-\frac{c_1}{x+1}+.......+{(-1)}^n\frac{c_n}{x+n}$

We have to calculate the the value of integration in order to find the required value of RHS.

${\int }_0^1\frac{y^{x-1}}{1}{(1-y)}^{n}dy$ ...... (using by parts)

$={\left[\frac{y^x}{x}{(1-y)}^n\right]}_0+n{\int }_0^1\frac{y^x}{x}{(1-y)}^{n-1}dy$

$=0+\frac{n}{x}{\int }_0^1{y}^x{(1-y)}^{n-1}dy$

Again by using by parts:-

$=\frac{n}{x}\cdot {[\frac{y^{x+1}}{x+1}\cdot {(1-y)}^{n-1}]}_0+{\int }_0^1(n-1){(1-y)}^{n-2}\frac{y^{x+1}}{x+1}$

$=\frac{n}{x}\cdot [0+\frac{n-1}{x+1}{\int }_0^1{y}^{x+1}{(1-y)}^{n-2}dy]$

$=\frac{n(n-1)}{x(x+1)}{\int }_0^1{y}^{x+1}{(1-y)}^{n-2}dy$

This integration by parts will go untill ${(1-y)}^{n-2}$becomes${(1-y)}^0$i.e. 1 and then the integration will stop.

So, final integration will be

${\int }_0^1{y}^{x+n-1}dy=\frac{y^{x+n}}{x+n}=\frac{1}{x+n}$

Thus, ${\int }_0^1{y}^{x-1}(1-y{)}^n=\frac{n(n-1)(n-2).....1}{x(x+1)(x+2)....(x+n)}=\frac{n!}{x(x+1)...(x+n)}$

Hence, the correct option is C.