Let α>0, be the smallest number such that the expansion of (x^(2/3)+2/x³)³⁰ has a term β x^(-α), β ∈ ℕ. Then α is equal to ___________.

Solution

Maths sir · 4 steps

Write the power of x in the general term, then pick the first r that makes it negative.

1. General term. The (r+1)th term is C(30, r)(x^(2/3))³⁰⁻ʳ(2/x³)ʳ = C(30, r) 2ʳ x^(2(30-r)/3 - 3r).

   Tᵣ₊₁ = C(30, r) 2ʳ x^((60 - 11r)/3), r = 0, 1, …, 30

2. Match the form. The coefficient C(30, r) 2ʳ is always a natural number, so any term works for β. We need -α = (60-11r)/3, that is α = (11r - 60)/3.
3. Smallest positive α. α > 0 needs 11r > 60, so r > 5.45, giving r ≥ 6. Since α grows with r, the smallest value comes from r = 6: α = (66 - 60)/3 = 2.
4. So. The term is C(30, 6) 2⁶ x⁻² = 38001600 x⁻², and α = 2.

The idea to keep. For any condition on a power of x, first combine every exponent into one expression in r. Then the question is only an inequality on r, with r a whole number from 0 to n.

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