Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Manipal MET2021MathematicsBinomial Theorem

The term independent of x in the expansion of (2 x^4- 1 x^2 )¹² is

Options

  1. A6920
  2. B7920
  3. C7900
  4. D3960

Correct answer

B. 7920

Step-by-step solution

The general term is aligned T_ r+1 & = ¹² C_r (2 x^4 )^r (- 1 x^2 )^ 12-r & = ¹² C_r(2)^r x^ 4 r (-1)^ 12-r (x)^ -2(12-r) aligned For term independent of x , we put aligned & 4 r-24+2 r=0 6 r=24 r=4 & aligned T₅ & = ¹² C₄(2)^4(-1)¹²⁻⁴ & = 12 11 10 9 4 3 2 1 (2)^4=7920 aligned aligned

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 Manipal MET PYQs