In the expansion of (∛2+1/∛3)ⁿ, n ∈ N, if the ratio of 15ᵗʰ term from the beginning to the 15ᵗʰ term from the end is 1/6, then the value of ⁿ C₃ is

Solution

Maths sir · 4 steps

The 15th term from the end is the 15th term of the reversed binomial, so the ratio is a power of 6.

1. Name the terms. Let a = 2^(1/3) and b = 3^(-1/3). The 15th term from the beginning of (a+b)ⁿ is T₁₅ = ⁿC₁₄ aⁿ⁻¹⁴ b¹⁴.
2. Term from the end. The 15th term from the end of (a+b)ⁿ is the 15th term from the beginning of (b+a)ⁿ, which is ⁿC₁₄ bⁿ⁻¹⁴ a¹⁴.
3. Take the ratio. The ⁿC₁₄ cancels, leaving aⁿ⁻²⁸/bⁿ⁻²⁸ = (a/b)ⁿ⁻²⁸. Here a/b = 2^(1/3) × 3^(1/3) = 6^(1/3).

   6^((n-28)/3) = 1/6 = 6⁻¹ ⇒ (n-28)/3 = -1 ⇒ n = 25

4. So. ²⁵C₃ = (25 × 24 × 23)/6 = 2300, which is option (b).

The idea to keep. The rth term from the end of (a+b)ⁿ is the rth term of (b+a)ⁿ. In a ratio of matching terms from each end, the binomial coefficients cancel, so you get (a/b)ⁿ⁻²⁽ʳ⁻¹⁾.

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