Question

If '$n$' is a positive integer and '$x$' is any non-zero number, then prove that
$C_0+C_1.\frac{x}{2}+C_2.\frac{x^2}{3}+.....+C_n.\frac{x^n}{n+1}=\frac{(1+x{)}^{n+1}-1}{(n+1)x}$

Answer 1

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.....C_n{x}^n$

$\frac{(1+x{)}^n}{(n+1)}=C_0+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+....C_n\frac{x^{n+1}}{n+1}+C$

Put $x=0LHS=\frac{\left(1\right)}{(n+1)}$

$RHS=C$

$\therefore C=\frac{1}{n+1}$

$\frac{(1+x{)}^{n+1}}{n+1}-C=C_{0}x+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+.....\frac{C_n{x}^{n+1}}{x+1}$

$\frac{\left[\right(1+x{)}^{n+1}-1]}{(n+1)}=x(C_0+\frac{C_{1}x}{2}+\frac{C_2{x}^2}{3}+....\frac{C_n{x}^n}{(n+1)})$

$\therefore C_0+C_1\frac{x}{2}+C_2\frac{x^2}{3}+....\frac{C_n{x}^n}{(n+1)}=\frac{\left(\right(1+x{)}^{n+1}-1)}{(n+1)x}$