Let aₙ = ∫₋₁ⁿ ( 1 + x/2 + x²/3 + ..... + xⁿ⁻¹/n )dx for every n ∈ N. Then the sum of all the elements of the set n ∈ N : aₙ ∈ (2, 30) is ____________.

Solution

Maths sir · 3 steps

Integrate term by term to get a clean formula for aₙ, then test n = 1, 2, 3, 4 one by one.

1. Integrate each term. The general term is xᵏ⁻¹/k, and its antiderivative is xᵏ/k². Put in the limits n and -1 for every k from 1 to n.

   aₙ = Σₖ₌₁ⁿ (nᵏ - (-1)ᵏ)/k²

2. Check small n. For n ≥ 1, every term in the sum is positive. Going from aₙ to aₙ₊₁, each old term grows and one more term is added, so aₙ increases with n. Compute values until aₙ crosses 30.

   n	Terms	aₙ	In (2, 30)?
   1	(1+1)/1	2	no, 2 is excluded
   2	3 + 3/4	3.75	yes
   3	4 + 2 + 28/9	≈ 9.11	yes
   4	5 + 15/4 + 65/9 + 255/16	≈ 31.91	no, above 30

3. So. Since aₙ keeps increasing, every n ≥ 4 gives more than 30. The set is {2, 3}, and the sum of its elements is 5.

The common slip: 6. Counting n = 1 gives 6, but a₁ = 2 and the interval (2, 30) is open, so 2 is not included.

The idea to keep. When the integrand is a sum of powers, integrate term by term to get a closed form. Then check whether the sequence is increasing, so you know where to stop testing values.

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