If lim _(n → ∞) (n+1)ᵏ⁻¹/nᵏ⁺¹[(n k+1)+(n k+2)+…+(n k+n)] ; =33 · lim _(n → ∞) 1/nᵏ⁺¹ ·[1ᵏ+2ᵏ+3ᵏ+…+nᵏ], then the integral value of k is equal to _____________

Solution

Maths sir · 5 steps

Turn both sides into plain numbers in k first, then the condition becomes a quadratic in k.

1. Add the bracket. There are n terms nk + r for r = 1 to n. Their sum is n · nk + (1 + 2 + … + n) = n² k + n(n+1)/2.
2. Left side limit. For large n, (n+1)ᵏ⁻¹ behaves like nᵏ⁻¹, and the sum behaves like n²(k + 1/2). The numerator is then about nᵏ⁺¹(k + 1/2), so the left side tends to k + 1/2.
3. Right side limit. As a limit of a sum, 1/nΣᵣ₌₁ⁿ (r/n)ᵏ → ∫₀¹ xᵏ dx = 1/(k+1). So the right side is 33/(k+1).

   (k + 1/2)(k+1) = 33 ⇒ 2k² + 3k - 65 = 0

4. Solve. The discriminant is 9 + 520 = 529 = 23², so k = (-3 ± 23)/4. This gives k = 5 or k = -13/2.
5. So. The only integer value is k = 5.

The idea to keep. Whenever you see 1/nᵏ⁺¹Σ rᵏ, read it as ∫₀¹ xᵏ dx = 1/(k+1). For the other side, keep only the highest power of n in the numerator and the denominator.

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