If the value of \(\left(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots . . \text { upto } \infty\right)^{\log _{(0.25)}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . \text { upto } \infty\right)}\) is I , then \(\mathrm{I}^{2}\) is equal to

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Sequences and Series

If the value of

(1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \dots .. upto \infty)^{l o g_{(0.25)} (\frac{1}{3} + \frac{1}{3 ^{2}} + \frac{1}{3 ^{3}} + \dots . upto \infty)}

is I , then

I^{2}

is equal to

Answer: 3

The series in the exponent is geometric with ratio

\frac{1}{3}

\frac{1}{3} + \frac{1}{3 ^{2}} + \frac{1}{3 ^{3}} + \dots = \frac{1/3}{1 - 1/3} = \frac{1}{2}, lo g_{0.25} \frac{1}{2} = \frac{ln \frac{1}{2}}{ln \frac{1}{4}} = \frac{1}{2}

In the base, the numerators after the leading

1

are

2, 6, 10, \dots

, an arithmetic progression with common difference

4

, so the

n

th of them is

4 n - 2

and

1 + n = 1 \sum \infty \frac{4 n - 2}{3 ^{n}} = 1 + 4 n = 1 \sum \infty \frac{n}{3 ^{n}} - 2 n = 1 \sum \infty \frac{1}{3 ^{n}} = 1 + 4 \cdot \frac{3}{4} - 2 \cdot \frac{1}{2} = 3

using

n \geq 1 \sum n x^{n} = \frac{x}{( 1 - x ) ^{2}}

x = \frac{1}{3}

. Hence

I = 3^{1/2}

and

I^{2} = 3

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer