Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET202423 May 2024Morning ShiftMathematicsBinomial TheoremActual

The independent term in the expansion of (1+x+2 x^2 ) ( 3 x^2 2 - 1 3 x )^9 is

Options

  1. A18 7
  2. B7 18
  3. C- 7 18
  4. D- 18 7

Correct answer

B. 7 18

Step-by-step solution

The general term in the expansion of ( 3 x^2 2 - 1 3 x )^9 is T_ r+1 = ^9 C_r ( 3 x^2 2 )^ 9-r ( -1 3 x )^r= ^9 C_r ( 3 2 )^ 9-r ( -1 3 )^r x^ 18-3 r Now, the general tem in the given expansion aligned & = (1+x+2 x^2 ) ( ^9 C_r ( 3 2 )^ 9-r ( -1 3 )^r x^ 18-3 r ) & = ^9 C_r ( 3 2 )^ 9-r ( -1 3 )^r x^ 18-3 r + ^9 C_r ( 3 2 )^ 9-r ( -1 3 )^r & + + ²⁹ C_r ( 3 2 )^ 9-r ( -1 3 )^r x^ 20-3 r aligned for independent of x, 18-3 r=0,19-3 r20-3 r=0 r=6 So, required term = ^9 C₆ ( 3 2 )⁹⁻⁶ ( -1 3 )^6= 7 18 .

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 AP EAMCET PYQs