Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
66 views
in Binomial Theorem by (115k points)
closed by
The middle term in the expansion of \(\rm\left(x^2-\frac{2}{x^3}\right)^{11}\), is:
1. 491\(\rm \frac{1}{x^{34}}\)
2. 1672\(\rm \frac{1}{x^{23}}\)
3. -14784\(\rm \frac{1}{x^3}\)
4. None of these.

1 Answer

0 votes
by (114k points)
selected by
 
Best answer
Correct Answer - Option 3 : -14784\(\rm \frac{1}{x^3}\)

Concept:

Binomial Expansion:

  • (a + b)n = Can b+ Can-1 b1 + Can-2 b2 + … + Can-r br + … + Cn-1 a1 bn-1 + Cn a0 bn, where C0, C1, …, Cn are the Binomial Coefficients defined as Cr = nCr = \(\rm \frac{n!}{r!(n-r)!}\).
  • The total number of terms in the expansion is n + 1.
  • The (r + 1)th term in the expansion is Tr+1 = Cr an-r br.
  • The central term is \(\rm T_{\tfrac{n}{2}}\) if n is even, and both \(\rm T_{\tfrac{n+1}{2}}\) and \(\rm T_{\tfrac{n+3}{2}}\) are the middle terms if n is odd.


Calculation:

In the given expression \(\rm\left(x^2-\frac{2}{x^3}\right)^{11}\), n = 11 (odd).

∴ There are two middle terms: \(\frac{11+1}{2}\) = 6th and \(\frac{11+3}{2}\) = 7th terms.

T6 = C5 \(\rm (x^2)^{11-5}\left(\frac{-2}{x^3}\right)^{5}\) = -14784\(\rm \frac{1}{x^3}\)

T7 = C6 \(\rm (x^2)^{11-6}\left(\frac{-2}{x^3}\right)^{6}\) = 29568\(\rm \frac{1}{x^8}\)

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.

...