JEE Main 2020 · 3 September · morning · Single correct · +4 / −1

If the number of integral terms in the expansion
of (3 1/2 + 5 1/8 )ⁿ is exactly 33, then the least value of n is :

  1. (a) 264
  2. (b) 256
  3. (c) 128
  4. (d) 248
Answer and solution

Answer (b) 256

Solution

Maths sir · 4 steps

A term is an integer only when both powers come out whole, so make r a multiple of 8 and n even.

  1. General term. The (r+1)th term is Tᵣ₊₁ = C(n, r) 3^((n-r)/2) 5^(r/8), for r = 0, 1, …, n.
  2. Conditions. For an integer, r/8 must be a whole number, so r is a multiple of 8. Also (n-r)/2 must be whole, so n - r is even. Every multiple of 8 is even, so this needs n to be even. If n were odd, there would be no integral terms at all.
  3. Count the terms. With n even, the allowed values are r = 0, 8, 16, … up to the largest multiple of 8 not exceeding n.

    Number of integral terms = ⌊ n/8 ⌋ + 1 = 33 ⇒ ⌊ n/8 ⌋ = 32

  4. So. This needs 256 ≤ n ≤ 263, with n even. The least such value is n = 256, which is option (b).

The idea to keep. For integral terms, list one divisibility condition for each power, then count the values of r that satisfy all of them. Remember that r = 0 also counts, which is where the +1 comes from.

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

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2020.