Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Find the sum of n terms of the series 1/1.2.3.4 + 1/2.3.4.5 + 1/3.4.5.6 +.........

 

Answers (1)

best_answer

Below you can find the answer: 

\\t_r=\frac{1}{r(r+1)(r+2)(r+3)}\\t_{r+1}=\frac{1}{(r+1)(r+2)(r+3)(r+4)}\\\frac{t_r}{t_{r+1}}=\frac{r+4}{r}\\r.t_r=(r+4)t_{r+1}\\r.t_r=(r+1)t_{r+1}+3t_{r+1} \\Putting \:\:r=1,2,3.......n-1\\adding\: \:we \:get,t_1-nt_n= 3(t_1+t_2+t_3.....t_n)\\or\:\:4t_1-nt_n=3(t_1+t_2+t_3.....t_n)=3S_n\\S_n=\frac{1}{3}(4t_1-nt_n)=\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}

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 Mains 2026 Exam City Slip LIVE: NTA session 2 test centre allotment soon; admit card date
anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 city slip soon; admit card, exam dates, here's how to prepare

Latest Question