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JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Sequences and Series

If

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}

at

x = \frac{1}{2}

is:

(A) $2 e^{2}$
(B) $\frac{1}{2} e^{2}$ ✓ Correct
(C)2 e
(D) $\frac{1}{2} e$

Answer: (B)

Solution

The coefficient of

x^{n}

is

\frac{n - 1}{n} = 1 - \frac{1}{n}

for

n \geq 2

, so

y = n \geq 2 \sum x^{n} - n \geq 2 \sum \frac{x ^{n}}{n} = \frac{x ^{2}}{1 - x} + ln (1 - x) + x .

At

x = \frac{1}{2}

,

y = \frac{1/4}{1/2} + ln \frac{1}{2} + \frac{1}{2} = 1 - ln 2

, so

1 + y = 2 - ln 2

and

e^{1 + y} = e^{2 - l n 2} = \frac{e ^{2}}{2} .

Thus

e^{1 + y} = \frac{1}{2} e^{2}

.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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