Let the coefficients of three consecutive terms Tᵣ, Tᵣ₊₁ and Tᵣ₊₂ in the binomial expansion of (a + b)¹² be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (∜3+∛4)¹². Then p + q is equal to:

Solution

Maths sir · 5 steps

Test the G.P. condition with ratios of consecutive coefficients; it forces n + 1 = 0, so p = 0.

1. Set up the G.P. The coefficients of Tᵣ, Tᵣ₊₁, Tᵣ₊₂ are ¹²Cᵣ₋₁, ¹²Cᵣ, ¹²Cᵣ₊₁. Three numbers are in G.P. exactly when the two consecutive ratios are equal. Each ratio uses ⁿCₖ/ⁿCₖ₋₁ = (n-k+1)/k.

   (n-r+1)/r = (n-r)/(r+1) ⇒ nr - r² + n + 1 = nr - r² ⇒ n + 1 = 0

2. No value of r. With n = 12 this says 13 = 0, which is false for every r. So no three consecutive binomial coefficients are in G.P., and p = 0.
3. General term. In (3^(1/4) + 4^(1/3))¹², Tₖ₊₁ = ¹²Cₖ 3^((12-k)/4) 4^(k/3). The term is rational only when k is a multiple of 3 and 12 - k is a multiple of 4. The second condition means k itself is a multiple of 4.
4. Rational terms. A multiple of both 3 and 4 in 0 ≤ k ≤ 12 is k = 0 or k = 12. These give 3³ = 27 and 4⁴ = 256, so q = 27 + 256 = 283.
5. So. p + q = 0 + 283 = 283, which is option (b).

The idea to keep. Consecutive binomial coefficients can be in A.P., but never in G.P.: equating the ratios always gives n + 1 = 0. For rational terms, make every fractional exponent in Tₖ₊₁ an integer.

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