If \(1+\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}+\frac{5-2\sqrt{6}}{18}+\frac{9\sqrt{3}-11\sqrt{2}}{36\sqrt{3}}+\frac{49-20\sqrt{6}}{180}+\ldots\) upto \(\infty=2+\left(\sqrt{\frac{b}{a}}+1\right)\log_{e}\left(\frac{a}{b}\right)\), where \(a\) and \(b\) are integers with \(\gcd(a,b)=1\), then \(11a+18b\) is equal to ____ .

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Sequences and Series

1 + \frac{3 - 2}{2 3} + \frac{5 - 2 6}{18} + \frac{9 3 - 11 2}{36 3} + \frac{49 - 20 6}{180} + \dots

upto

\infty = 2 + (\frac{b}{a} + 1) lo g_{e} (\frac{a}{b})

, where

a

and

b

are integers with

g cd (a, b) = 1

, then

11 a + 18 b

is equal to ____ .

Answer: 76

The series has terms of the form

\frac{1}{n} (\frac{x}{3})^{n}

-type coefficients after substituting

t = \frac{3 - 2}{2 3}

(equivalently

x = 3 - 23

scaled by

\frac{1}{3}

) and recognising the numerators as

1, t, t^{2}, t^{3}, \dots

built from binomial-type coefficients

(2 n)

. Summing the resulting series

S = 2 - (1 - \frac{1}{t}) - lo g (1 - t) (\frac{1}{t} - 1)

and simplifying with

t = \frac{3 - 2 3}{3}

leads, after rationalising, to

S = 2 + (\frac{3}{2} + 1) lo g_{e} (\frac{2}{3})

a = 2, b = 3

(and indeed

g cd (2, 3) = 1

11 a + 18 b = 11 (2) + 18 (3) = 22 + 54 = 76

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Integer answer