Let the coefficients of x⁻¹ and x⁻³ in the expansion of ( 2x^(1/5) - 1/x^(1/5) )¹⁵,x > 0, be m and n respectively. If r is a positive integer such that mn² = ¹⁵Cᵣ . 2ʳ, then the value of r is equal to __________.

Solution

Maths sir · 5 steps

Write the general term, find which k gives each power of x, then match mn² to ¹⁵Cᵣ · 2ʳ.

1. General term. The (k+1)th term is ¹⁵Cₖ (2x^(1/5))¹⁵⁻ᵏ(-x^(-1/5))ᵏ. Collect the powers of 2, of -1 and of x.

   Tₖ₊₁ = ¹⁵Cₖ 2¹⁵⁻ᵏ (-1)ᵏ x^((15-2k)/5)

2. Coefficient m. For x⁻¹ you need (15-2k)/5 = -1, so k = 10. Then m = ¹⁵C₁₀· 2⁵·(-1)¹⁰ = ¹⁵C₁₀· 2⁵.
3. Coefficient n. For x⁻³ you need (15-2k)/5 = -3, so k = 15. Then n = ¹⁵C₁₅· 2⁰·(-1)¹⁵ = -1, so n² = 1.
4. Match the form. mn² = ¹⁵C₁₀· 2⁵ = ¹⁵C₅· 2⁵, using ¹⁵C₁₀ = ¹⁵C₅. Now the binomial coefficient and the power of 2 both carry the same index, 5.
5. So. ¹⁵Cᵣ· 2ʳ = ¹⁵C₅· 2⁵ gives r = 5.

The common slip: 10. Matching only the ¹⁵C₁₀ part gives 10, but then the power of 2 would need to be 2¹⁰, not 2⁵. Rewrite ¹⁵C₁₀ as ¹⁵C₅ so both parts match with r = 5.

The idea to keep. For any power of x, set the exponent in the general term equal to it and solve for k. When matching a form like ⁿCᵣ· aʳ, use ⁿCᵣ = ⁿCₙ₋ᵣ so both indices agree.

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