JEE Main 2023 · 31 January · evening · Numerical answer · +4 / −0

The coefficient of x⁻⁶, in the
expansion of ((4 x)/5+5/(2 x²))⁹, is

Answer and solution

Answer 5040

Solution

Maths sir · 4 steps

Write the general term, match the power of x to -6, and that fixes r.

  1. General term. In (a+b)⁹ the term Tᵣ₊₁ is C(9, r)a⁹⁻ʳbʳ. Here a = 4x/5 and b = 5/2x².

    Tᵣ₊₁ = C(9, r)(4/5)⁹⁻ʳ(5/2)ʳ x⁹⁻³ʳ

  2. Find r. The power of x is (9-r) - 2r = 9 - 3r. Setting 9 - 3r = -6 gives r = 5, so the term we need is T₆.
  3. Coefficient. It is C(9, 5)(4/5)⁴(5/2)⁵ = 126 × 256/625 × 3125/32. Pair them up: 256/32 = 8 and 3125/625 = 5.
  4. So. The coefficient is 126 × 8 × 5 = 5040.

The idea to keep. For a specific power of x, first collect the power of x in Tᵣ₊₁ as a function of r, solve for r, and only then work out the numbers.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2023.