Mathematics

Problem

0 < x < 1

and

y = \frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \dots

, then the value of

e^{1+y}

x = \frac{1}{2}

is (a)

\frac{1}{2}e^2

(b)

2e

(c)

2e^2

(d)

\frac{1}{2}\sqrt{e}