Question

Prove that $1-{}^n{C}_1\frac{1+x}{1+nx}+{}^n{C}_2\frac{1+2x}{1+n{x}^2}-{}^n{C}_3\frac{1+3x}{1+n{x}^3}+.....(n+1)$ terms = 0

Answer 1

Put $\frac{1}{1+nx}=t$ and split into two .
$(C_0-C_{1}t+C_2{t}^2-C_3{t}^3+...)+x(-C_{1}t+2C_2{t}^2-3c_3{t}^3+....)=A+xB$
$A=C_0-C_{1}t+C_2{t}^2+-C_3{t}^3+....$ ......(1)
$(1-t{)}^n={(1-\frac{1}{1+nx})}^n={\left(\frac{nx}{1+nx}\right)}^n$
In order to get B , we differentiate (1)
$-n(1-t{)}^{n-1}=-C_1+2C_{2}t-3C_3{t}^2+....$
Multiply both sides by t
$-n(1-t{)}^{n-1}=-C_1+2C_{2}t-3C_3{t}^2+....=B$
$\therefore B=-n\frac{1}{1+nx}{[1-\frac{1}{1+nx}]}^{n-1}$
$=\frac{-n}{1+nx}{\left(\frac{nx}{1+nx}\right)}^{n-1}$ ....(2)
$\therefore A+xB$
$={\left(\frac{nx}{1+nx}\right)}^n-\frac{nx}{1+nx}{\left(\frac{nx}{1+nx}\right)}^{n-1}$
$={\left(\frac{nx}{1+nx}\right)}^n-{\left(\frac{nx}{1+nx}\right)}^n=0$