The number of positive integers k such that the constant term in the binomial expansion of ( 2x³ + 3/xᵏ )¹², x ≠ 0 is 2⁸ . l, where l is an odd integer, is ______________.

Solution

Maths sir · 4 steps

First find every k that gives a constant term, then count the factors of 2 in each one.

1. General term. Tᵣ₊₁ = C(12, r)(2x³)¹²⁻ʳ(3/xᵏ)ʳ = C(12, r) 2¹²⁻ʳ 3ʳ x³⁶⁻³ʳ⁻ᵏʳ.
2. Constant term. Set the power of x to zero: 36 = r(3+k), so 3+k must divide 36. Since k ≥ 1, 3+k is one of 4, 6, 9, 12, 18, 36. That gives r = 9, 6, 4, 3, 2, 1, all within 0 ≤ r ≤ 12.
3. Count powers of 2. The factor 3ʳ is odd, so the power of 2 comes only from C(12, r) and from 2¹²⁻ʳ. We need exactly 2⁸, with an odd number left over.

   r	k	C(12, r)	Power of 2 in the term
   9	1	220 = 2² × 55	2 + 3 = 5
   6	3	924 = 2² × 231	2 + 6 = 8
   4	6	495, odd	0 + 8 = 8
   3	9	220 = 2² × 55	2 + 9 = 11
   2	15	66 = 2 × 33	1 + 10 = 11
   1	33	12 = 2² × 3	2 + 11 = 13

4. So. Only k = 3 and k = 6 give a constant term of the form 2⁸ · l with l odd. The number of such k is 2.

The common slip: 6. Counting every k that gives a constant term, and skipping the power-of-2 check, gives 6.

The idea to keep. For a condition like '2⁸ times an odd number', track the exact power of 2 in each piece, and do not forget the 2s hiding inside C(n, r).

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