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JEE MainMathematicsBinomial Theorem

Let P(x) = _ k=0 ⁴⁰ (3+x)^ 40-k (2+x)^k . The coefficient of x²⁰ in the polynomial P(x) is

Options

  1. A⁴⁰C₂₀(3²⁰ - 2²⁰)
  2. B⁴¹C₂₀(3²¹ + 2²¹)
  3. C⁴¹C₂₀(3²¹ - 2²¹)
  4. D⁴¹C₂₀(3²⁰ - 2²⁰)

Correct answer

C. ⁴¹C₂₀(3²¹ - 2²¹)

Step-by-step solution

The given polynomial is P(x) = _ k=0 ⁴⁰ (3+x)^ 40-k (2+x)^k . This is a geometric progression with the first term a = (3+x)⁴⁰ , common ratio r = 2+x 3+x , and the number of terms n = 41 . Using the sum formula for a G.P., S_n = a 1 - r^n 1 - r , we get: P(x) = (3+x)⁴⁰ 1 - ( 2+x 3+x )⁴¹ 1 - 2+x 3+x P(x) = (3+x)⁴⁰ (3+x)⁴¹ - (2+x)⁴¹ (3+x)⁴¹ (3+x) - (2+x) 3+x P(x) = (3+x)⁴¹ - (2+x)⁴¹ 1 = (3+x)⁴¹ - (2+x)⁴¹ We need to find the coefficient of x²⁰ in P(x) . The coefficient of x²⁰ in (3+x)⁴¹ is ⁴¹C₂₀ 3⁴¹⁻²⁰ = ⁴¹C₂₀ 3²¹ . Th

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