Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
KCET2016MathematicsBinomial Theorem

The 11th term in the expansion of ( (x+ 1 x )¹⁴ ) is

Options

  1. A( 999 x )
  2. B( 1001 x )
  3. C( i )
  4. D( x 1001 )

Correct answer

B. ( 1001 x )

Step-by-step solution

Given expression, ( (x+ 1 x )¹⁴ ) We know that nth term is given by [ array l T_ r+1 = n r a^ n-r b^ r So, T₁₁= 14 10 ( ⁴ ) ( 1 )¹⁰ = 14 10 ( ⁴ ) ( 1 ⁵ ) = 14 10 = 1001 array ]

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If n 13 , n 14 and n 15 are in arithmetic progression, then the positive integer value of ' n ' can be 2026If the coefficients of x^2 and x^3 in the expansion of (3 + kx)^9 are equal, then the value of ' k ' is 2026The remainder when 7¹⁰³ is divided by 25 is 2026_ r=1 ¹⁵ r^2 ( ¹⁵ C_r 15 r-1 )= 20251 81^ n - ^ 2 n C ₁ 10 81^ n + ^ 2 n C ₂ 10^2 81^ n - + 10^ 2 n 81^ n = 2025If x is positive real number and the first negative term in the expansion of (1+ x )^ 27 / 5 is t _ k then k = 2025In the binomial expansion of (p-q)¹⁴ , if the sum of 7^ th term and 8^ th term is zero, then p+q p-q = 2025The numerically greatest term in the expansion of (x+3 y)¹³ , when x= 1 2 and y= 1 3 is 2025 Full Binomial Theorem list All KCET PYQs