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
287 views
in Binomial Theorem by (44.7k points)
closed by

Find the two middle terms in the expansion of \(\Big(x^4- \frac{1}{x^3}\Big)^{11}\)

1 Answer

+1 vote
by (42.5k points)
selected by
 
Best answer

For \(\Big(x^4- \frac{1}{x^3}\Big)^{11}\)

a = x4, b = \(\frac{-1}{x^4}\) and n = 11

As n is odd, there are two middle terms i.e

II. \(\Big(\frac{n+1}{2}\Big)^{th}\) and II. \(\Big(\frac{n+3}{2}\Big)^{th}\) 

General term tr+1 is given by,

tr+1 = (rn) an-rbr

I. The first middle term is \(\Big(\frac{n+1}{2}\Big)^{th}\) \(\Big(\frac{11+1}{2}\Big)^{th}\) =   \(\Big(\frac{12}{2}\Big)^{th}\) = (6)th

Therefore, for the 6th middle term, r = 5 

Therefore, the first middle term is

t6 = t5+1

= -462.x9

II. The second middle term is  \(\Big(\frac{n+3}{2}\Big)^{th}\) \(\Big(\frac{11+3}{2}\Big)^{th}\) =   \(\Big(\frac{14}{2}\Big)^{th}\) = (7)th

Therefore, for the 7th middle term, r = 6 

Therefore, the second middle term is

t7 = t6+1

= 462.x2

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.

...