Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Sir...solve this ques 16 The sum of n terms of the series 11 1.2 3 2.3 32 a) 211-1 3n 3.4 33 4.5 34

Answers (1)

best_answer

@Mansi

This is telescoping series.

A sum in which subsequent terms cancel each other, leaving only initial and final terms. For example,

sum_(i=1)^(n-1)(a_i-a_(i+1))

=(a_1-a_2)+(a_2-a_3)+...+(a_(n-2)-a_(n-1))+(a_(n-1)-a_n)

=(a_1-a_n)

is a telescoping sum.

So, general term of the series

t_{n}=\frac{2n+3}{n(n+1)}*\frac{1}{3^{n}}=(\frac{3}{n}-\frac{1}{n+1})*\frac{1}{3^{n}}=\frac{1}{n*3^{n-1}}-\frac{1}{(n+1)3^{n}}

\sum t_{n}=1-\frac{1}{n+1}*\frac{1}{3^{n}}

 

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question