Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.....C_n{x}^n$then
$3C_0+3^2\frac{C_1}{2}$ + $\frac{3^3{C}_2}{3}+\frac{3^4{C}_3}{4}$ + ....+ $\frac{3^{n+1}{C}_n}{n+1}$ = $\frac{4^{n+1}-1}{n+1}$

• A. True
• B. False

Answer 1

The correct option is A True

$(1+x{)}^n=c_0+c_{1}x+c_2{x}^2+\dots c_n{x}^n$

$\text{on integrating both sides with respect}$

$\text{to}x\text{we get,}$

${\int }_0^x(1+x{)}^{n}dx={\int }_0^x(c_0+c_{1}x+c_2{x}^2+\dots -c_n{x}^n)dx$

$\Rightarrow {\left[\frac{(1+x{)}^{n+1}}{n+1}\right]}_0^x=c_{0}x+c_1\frac{x^2}{2}+c_2\frac{x^3}{3}+\dots +c_n\frac{x^{n+1}}{n+1}$

$\Rightarrow \frac{(1+x{)}^{n+1}-1}{n+1}=(c_{0}x+c_1\frac{x^2}{2}+c_2\frac{x^3}{3}+\cdots +c_n\frac{x^{n+1}}{n+1})$

at $x=3$ we get,

$\Rightarrow 3C_0+3^2\frac{C_1}{2}+\frac{3^3{C}_2}{3}+\dots +\frac{C_n\left(3^{n+1}\right)}{h+1}=\frac{4^{n+1}-1}{n+1}$

So It is true