Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET201920 Apr 2019Morning ShiftMathematicsBinomial TheoremActual

If ( (1-p x)⁻¹ (1-q x) =a₀+a₁ x+a₂ x^2+a₃ x^3+ ), then (a_n= )

Options

  1. A( p^ n+1 -q^ n+1 q-p )
  2. B( p^ n+1 -q^ n+1 p-q )
  3. C( p^n-q^n q-p )
  4. D( p^n-q^n p-q )

Correct answer

B. ( p^ n+1 -q^ n+1 p-q )

Step-by-step solution

Since, ( aligned & (1-p x)⁻¹ (1-q x) =a₀+a₁ x+a₂ x^2+a₃ x^3+ & (1-p x)⁻¹=1+p x+p^2 x^2+p^3 x^3+ +p^n x^n+ aligned ) and ((1-q x)⁻¹=1+q x+q^2 x^2+q^3 x^3+ +q^n x^n+ ) Now, coefficient of (x^n ) in the expansion of ( aligned & (1-p x)⁻¹(1-q x)⁻¹ & =p^n+p^ n-1 q+p^ n-2 q^2+p^ n-3 q^3+ +q^n aligned ) ( aligned & a_n= p^n (1- ( q p )^ n+1 ) 1- q p = p^n (p^ n+1 -q^ n+1 ) p (p-q) p^ n+1 & = p^ n+1 -q^ n+1 p-q & So, a_n= p^ n+1 -q^ n+1 p-q aligned ) Hence, option (2) is correct.

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