Question

The value of $1-\frac{n}{1}\frac{(1+x)}{(1+nx)}+\frac{n(n-1)}{1.2}\frac{(1+2x)}{(1+nx{)}^2}-\frac{n(n-1)(n-2)}{\mathrm{1.2.3}}\frac{(1+3x)}{(1+nx{)}^3}+......$ is:

• A. $2$
• B. $1$
• C. $0$
• D. $3$

Answer 1

The correct option is C $0$

$S={\sum }_{r=0}^n{{}^{n}C}_r.\frac{1+rx}{{(1+nx)}^r}{(-1)}^r$

$={(-1)}^r{\sum }_{r=0}^n{{}^{n}C}_r\frac{1}{{(1+nx)}^r}+{(-1)}^r{\sum }_{r=0}^{n}r{{}^{n}C}_r\frac{1}{{(1+nx)}^r}$

$={\sum }_{r=0}^n{{}^{n}C}_r{\left(\frac{-1}{(1+nx)}\right)}^r+x{(-1)}^r{\sum }_{r=1}^{n}n{{}^{n-1}C}_{r-1}\frac{1}{{(1+nx)}^r}$

$={(1-\frac{1}{(1+nx)})}^n+nx.{\sum }_{r=1}^n{{}^{n-1}C}_{r-1}{\left(\frac{-1}{(1+nx)}\right)}^r$

$={\left(\frac{nx}{1+nx}\right)}^n+nx\left(\frac{-1}{1+nx}\right){\sum }_{r=1}^n{{}^{n-1}C}_{r-1}{\left(\frac{-1}{(1+nx)}\right)}^r$

$={\left(\frac{nx}{1+nx}\right)}^n+\left(\frac{-nx}{1+nx}\right)\left[{(1-\frac{1}{1+nx})}^{n-1}\right]$

$={\left(\frac{nx}{1+nx}\right)}^n-{\left(\frac{nx}{1+nx}\right)}^n$

$=0.$

Hence the answer is $0.$