Sum of the series 1212+13-14122+132+16123+133……is

=12 12-1222+1233-...... +1213-1322+1333-..... =12 log 1+12+12log 1+13 =12 log 32 43=12 log 2

Sum of the series 1212+13-14122+132+16123+133……is

=12 12-1222+1233-...... +1213-1322+1333-..... =12 log 1+12+12log 1+13 =12 log 32 43=12 log 2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Sum of the series $\frac{1}{2} (\frac{1}{2} + \frac{1}{3}) - \frac{1}{4} (\frac{1}{2^{2}} + \frac{1}{3^{2}}) + \frac{1}{6} (\frac{1}{2^{3}} + \frac{1}{3^{3}}) \dots \dots$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\log_{e} 3$

$\frac{1}{2} \log_{e} 2$

$\frac{1}{2} \log_{e} \frac{3}{2}$

$\log_{e} \frac{2}{3}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{rcl} = & \frac{1}{2} [\frac{1}{2} - \frac{{(\frac{1}{2})}^{2}}{2} + \frac{{(\frac{1}{2})}^{3}}{3} - . . . . . .] \\ + \frac{1}{2} (\frac{1}{3} - \frac{{(\frac{1}{3})}^{2}}{2} + \frac{{(\frac{1}{3})}^{3}}{3} - . . . . .) \\ = & \frac{1}{2} [\log (1 + \frac{1}{2})] + \frac{1}{2} [\log (1 + \frac{1}{3})] \\ = & \frac{1}{2} [\log \frac{3}{2} \frac{4}{3}] = \frac{1}{2} \log 2 \end{array}$