Let for the 9ᵗʰ term in the binomial expansion of (3+6 x)ⁿ, in the increasing powers of 6 x, to be the greatest for x=3/2, the least value of n is n₀. If k is the ratio of the coefficient of x⁶ to the coefficient of x³, then k+n₀ is equal to :

Solution

Maths sir · 4 steps

Put x = 3/2 so 6x = 9, then trap T₉ between its two neighbours with the term ratio.

1. Term ratio. At x = 3/2 the expansion is (3+9)ⁿ, and every term is positive. With Tᵣ₊₁ = C(n, r) 3ⁿ⁻ʳ 9ʳ, the ratio of consecutive terms is Tᵣ₊₁/Tᵣ = (n-r+1)/r × 9/3 = 3(n-r+1)/r.
2. Trap the 9th term. Use T₉ ≥ T₈ (put r = 8): 3(n-7)/8 ≥ 1, so n ≥ 9.67. Then use T₁₀ ≤ T₉ (put r = 9): 3(n-8)/9 ≤ 1, so n ≤ 11. Hence n = 10 or 11, and the least value is n₀ = 10.

   k = (C(10, 6) 3⁴ 6⁶)/(C(10, 3) 3⁷ 6³) = 210/120 × 6³/3³ = 7/4 × 8 = 14

3. Ratio of coefficients. In (3+6x)¹⁰ the coefficient of xʳ is C(10, r) 3¹⁰⁻ʳ 6ʳ. Divide the coefficient of x⁶ by the coefficient of x³ as shown above. The powers of 3 and 6 reduce to 2³ = 8, which gives k = 14.
4. So. k + n₀ = 14 + 10 = 24.

The idea to keep. For the greatest term, find Tᵣ₊₁/Tᵣ with the actual value of x put in. Then force the given term to be at least as big as both of its neighbours.

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