Question

Let $(1+x{)}^n=1+nx+\frac{n(n-1)x^2}{2}+\cdots \left|x\right|<1,nϵZ$ Then the sum of the series
$3+\frac{8}{3}+\frac{80}{3^3}+\frac{240}{3^4}\cdots$ is

• A. 9
• B. 27
• C. 12
• D. 101

Answer 1

The correct option is B 27

$(1+x{)}^n=3+\frac{8}{3}+\frac{80}{3^3}+\frac{240}{3^4}+\cdots .=1+nx+\frac{n(n-1)x^2}{2}+\cdots ie1+3\left(\frac{2}{3}\right)+6{\left(\frac{2}{3}\right)}^2+15{\left(\frac{2}{3}\right)}^4+\cdots$ .
On comparison, $n=-3$ and $x=-\frac{2}{3}\therefore {(1-\frac{2}{3})}^{-3}=27$