Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The middle term in the expansion of <img alt="\left(x+\frac{1}{x}\right)^{2 n}=k \frac{1 \cdot 3 \cdot 5 \cdot-(2 n-1) 2^n}{n!}" src="/latex-image/?%5Cleft%28x&plus;%5Cfrac%7B1%7D%7Bx%7D%5Crigh

The middle term in the expansion of \left(x+\frac{1}{x}\right)^{2 n}=k \frac{1 \cdot 3 \cdot 5 \cdot-(2 n-1) 2^n}{n!}, where k=

Option: 1

n


Option: 2

(2 n+1)


Option: 3

1

 


Option: 4

2


Answers (1)

best_answer

As 2 n is even, the middle term is T_{n+1}

\begin{aligned} T_{n+1} & ={ }^{2 n} C_n x^n\left(\frac{1}{x}\right)^{2 n-n} \\ & ={ }^{2 n} C_n=\frac{(2 n) !}{n ! n !} \\ & =\frac{(2 n)(2 n-1)(2 n-2) \ldots . .4 .3 .2 .1}{n ! n !} \end{aligned}\\ $$ $$ \begin{aligned} & =\frac{(2 n .2 n-2.2 n-4 \ldots . .4 .2)(2 n-1.2 n-3 \ldots .3 .1)}{n ! n !} \\ & =2^n \frac{(n . n-1 . n-2 \ldots . .1)(1.3 .5 \ldots \ldots-(2 n-1))}{n ! n !} \\ & =\frac{2^n(1.3 .5 \ldots . .(2 n-1))}{n !} \end{aligned}

Hence k=1

Posted by

shivangi.bhatnagar

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 2025: NIT Andhra Pradesh cut-offs drop for CSE, ECE, other BTech programmes this year; closing rank
anam-khan

JEE Main 2025 Live: Exam city slip soon at jeemain.nta.nic.in; admit card date, cut-off updates
anam-khan

BTech Admissions 2025: GUJCET registration ends on December 31; JEE Main, AEEE, WBJEE, SRMJEEE exam date

Latest Question