Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Highly selective Backlog Qs for JEE MainMathematicsBinomial Theorem

If the middle term in the expansion of (1+x)^ 2 n is the greatest term, then x lies in the interval

Options

  1. A( n n+1 , n+1 n )
  2. B( n+1 n , n n+1 )
  3. C(n-2, n)
  4. D(n-1, n)

Correct answer

A. ( n n+1 , n+1 n )

Step-by-step solution

In the expansion of (1+x)^ 2 n , middle term is ^ 2 n C _n x^n . Since, middle term is the greatest term. ^ 2 n C _n x^n> ^ 2 n C _ n-1 x^ n-1 and ^ 2 n C _r x^n> ^ 2 n C _ n+1 x^ n+1 aligned & x> ^ 2 n C _ n-1 ^ 2 n C _n and x < ^ 2 n C _n ^ 2 n C _ n+1 & Hence, x ( ^ 2 n C _n ^ 2 n C _ n+1 , ^ 2 n C _n ^ 2 n C _ n+1 ) aligned Hence, x ( ^ 2 n C _n ^ 2 n C _ n+1 , ^ 2 n C _n ^ 2 n C _ n+1 ) aligned & x ( array c (2 n) ! (n-1) !(2 n-n+1) n !(2 n-n) ! (2 n) ! (2 n) ! n !(2 n-n) ! (n-1) !(2 n-n+1) (2 n) ! array ) & x

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If 26 ( 2^3 3 12 2 + 2^5 5 12 4 + 2^7 7 12 6 + + 2¹³ 13 12 12 ) = 3¹³ - , then is equal to: 2026If (1 - x^3)¹⁰ = _ r=0 ¹⁰ a_r x^r (1-x)^ 30-2r , then 9a₉ a₁₀ is equal to __________. 2026If the coefficients of the middle terms in the binomial expansions of (1 + x)²⁶ and (1 - x)²⁸ , 0 , are equal, then the value of is: 2026The coefficient of x^2 in the expansion of (2x^2 + 1 x )¹⁰ , x 0 , is : 2026If the sum of the coefficients of x^7 and x¹⁴ in the expansion of ( 1 x^3 - x^4 )^n , x 0 , is zero, then the value of n is __________. 2026In the expansion of (9x- 1 3 x )¹⁸ , x>0 , if the term independent of x is (221)k , then k is equal to: 2026Let the smallest value of k N , for which the coefficient of x^3 in (1+x)^3 + (1+x)^4 + (1+x)^5 + + (1+x)⁹⁹ + (1+kx)¹⁰⁰ , x 0 , is (43n + 101 4 ) (¹⁰⁰C₃ ) for some n N , be p . The 2026If for 3 r 30 , 30 30-r + 3 30 31-r + 3 30 32-r + 30 33-r = m r , then m equals: 2026 Full Binomial Theorem list All Highly selective Backlog Qs for JEE Main PYQs