If 1 + 3 - 2 2 3 + 5 - 2 6 18 + 9 3 - 11 2 36 3 + 49 - 20 6 180 + ? upto = 2 + ( b a + 1 ) log e ( a b ) , where a and b are integer with g c d ( a , b ) = 1 , then 11 a + 18 b is equal to _______. [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If $1 + \frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{3}} + \frac{5 - 2 \sqrt{6}}{18} + \frac{9 \sqrt{3} - 11 \sqrt{2}}{36 \sqrt{3}} + \frac{49 - 20 \sqrt{6}}{180} + \dots$ upto $\infty = 2 + (\sqrt{\frac{b}{a}} + 1) \log_{e} (\frac{a}{b}),$ where $a$ and $b$ are integer with $g c d (a, b) = 1,$ then $11 a + 18 b$ is equal to _______. [2024]

Ans.
(76)

Let $S = 1 + \frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{3}} + \frac{5 - 2 \sqrt{6}}{18} + \frac{9 \sqrt{3} - 11 \sqrt{2}}{36 \sqrt{3}} + \dots \infty$

$= 1 + \frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{3}} + \frac{{(\sqrt{3} - \sqrt{2})}^{2}}{{(\sqrt{3})}^{2} \cdot 6} + \frac{{(\sqrt{3} - \sqrt{2})}^{3}}{{(\sqrt{3})}^{3} \cdot 12} + \dots \infty$

$= 1 + \frac{t}{2} + \frac{t^{2}}{6} + \frac{t^{3}}{12} + \dots \infty$ $[where t = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}}]$

$= 1 + t (1 - \frac{1}{2}) + t^{2} (\frac{1}{2} - \frac{1}{3}) + t^{3} (\frac{1}{3} - \frac{1}{4}) + \dots \infty$

$= (t + \frac{t^{2}}{2} + \dots) - \frac{1}{t} (t + \frac{t^{2}}{2} + \frac{t^{3}}{3} + \dots) + 2$

$= 2 + (\frac{1}{t} - 1) \log (1 - t)$

$= 2 + (\frac{\sqrt{3}}{\sqrt{3} - \sqrt{2}} - 1) \log (1 - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}})$

$= 2 + (\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}) \log (\frac{\sqrt{2}}{\sqrt{3}}) = 2 + (\frac{\sqrt{6} + 2}{1}) \frac{1}{2} \log \frac{2}{3}$

$= 2 + (\sqrt{\frac{3}{2}} + 1) \log \frac{2}{3}$

$\Rightarrow$ $a = 2$ and $b = 3$

$∴$ $11 a + 18 b = 22 + 54 = 76$

Want Us to Email you About Special Offers & Updates?