LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
17 views
in Binomial Theorem by (54.3k points)
closed by
Middle term in the expansion \(\rm (x-\frac2x)^{16}\) is
1. \(\rm ^{16}C_9 \times 2^8\)
2. \(\rm ^{16}C_9 \times 2^9\)
3. \(\rm ^{16}C_8 \times 2^8\)
4. \(\rm ^{16}C_8 \times 2^9\)

1 Answer

0 votes
by (30.0k points)
selected by
 
Best answer
Correct Answer - Option 3 : \(\rm ^{16}C_8 \times 2^8\)

Concept:

The general term in the expansion of (a - b)n is given by, \(\rm T_{r+1}=(-1)^r \times ^nC_ra^{n-r}b^r\)

  • When n is even, then the middle term = \(\rm (\frac n 2+1)\)th term.
  • When n is odd, then the middle term = \(\rm \frac 12(n+1)\)th term and \(\rm \frac 12(n+3)\)th term.

Calculation:

The general term in the expansion of \(\rm (x-\frac2x)^{16}\)is given by, 

\(\rm T_{r+1}=(-1)^r\times^{16}C_r\times x^{16-r}\times(\frac2x)^r\)

\(\rm ⇒ (-1)^r\times ^{16}C_r\times x^{16-2r}\times 2^{r}\)

 

Here, n is 16 (i.e., even), so there will be one middle terms

 \(\rm (\frac n2+1)\)th term = (8 + 1)th term = 9th term

∴ \(\rm T_{9}=T_{8+1}\) \(\rm =(-1)^8\times^{16}C_8\times x^{16-16} \times 2^8\)

⇒ \(\rm ^{16}C_8 \times 2^8\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...