lim n → ∞ ((1²-1)(n-1)+(2²-2)(n-2)+…+((n-1)²-(n-1)) · 1)/((1³+2³+… …+n³)-(1²+2²+… …+n²)) is equal to :

Solution

Maths sir · 4 steps

Write the general term, keep only the n⁴ part on top and bottom, and the limit is their ratio.

1. General term. The r-th term of the numerator is (r² - r)(n - r) for r = 1 to n-1. Expanded, it is (n+1)r² - r³ - nr.
2. Numerator. Use Σ r² ≈ n³/3 and Σ r³ ≈ n⁴/4. Then (n+1)Σ r² ≈ n⁴/3 and Σ r³ ≈ n⁴/4. The nr part gives only n³ terms, so the numerator behaves like n⁴/3 - n⁴/4 = n⁴/12.
3. Denominator. Σ r³ = n²(n+1)²/4 ≈ n⁴/4, while Σ r² is only of order n³. So the denominator behaves like n⁴/4.

   (n(n+1)(n-1)(n-2)/12)/(n(n+1)(n-1)(3n+2)/12) = (n-2)/(3n+2) → 1/3

4. So. The exact sums factor as shown above, and both routes give the limit (1/12)/(1/4) = 1/3, which is option (d).

The idea to keep. For a ratio of sums as n → ∞, find the highest power of n on top and bottom. Only those leading coefficients decide the limit; everything of lower degree vanishes.

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