Question

If $0<x<1$ and $y=\frac{1}{2}{x}^2+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+\cdots ,$ then the value of $e^{1+y}$ at $x=\frac{1}{2}$ is

• A. $2e$
• B. $\frac{1}{2}{e}^2$
• C. $2e^2$
• D. $\frac{1}{2}\sqrt{e}$

Answer 1

The correct option is B $\frac{1}{2}{e}^2$

$y=\frac{1}{2}{x}^2+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+\cdots$
$=(1-\frac{1}{2})x^2+(1-\frac{1}{3})x^3+(1-\frac{1}{4})x^4+\cdots$
$=(x^2+x^3+x^4+\cdots )+(-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\cdots )$
$=\frac{x^2}{1-x}+x+(-\frac{x}{1}-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\cdots )$
$y=\frac{x}{1-x}+\ln (1-x)$
$y+1=\frac{1}{1-x}+\ln (1-x)$
$e^{y+1}=e^{\frac{1}{1-x}+\ln (1-x)}=e^{\frac{1}{1-x}}\times e^{\ln (1-x)}=(1-x)e^{\frac{1}{1-x}}$
$\therefore$ at $x=\frac{1}{2},y=\frac{1}{2}{e}^2$