Question

$$ \lim_{n \to \infty} \frac{(1^2 - 1)(n-1) + (2^2 - 2)(n-2) + \ldots + ((n-1)^2 - (n-1))}{(1^3 + 2^3 + \ldots + n^3) - (1^2 + 2^2 + \ldots + n^2)} $$ is equal to: 

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

The Correct Option is B

The given problem requires evaluating the limit as \( n \to \infty \) of the expression:

\[\lim_{n \to \infty} \frac{(1^2 - 1)(n-1) + (2^2 - 2)(n-2) + \ldots + ((n-1)^2 - (n-1))}{(1^3 + 2^3 + \ldots + n^3) - (1^2 + 2^2 + \ldots + n^2)}\]

This requires simplifying both the numerator and the denominator separately before taking the limit as \( n \to \infty \).

1. Consider the numerator: 

\[\sum_{k=1}^{n-1} \left( k^2 - k \right)(n-k)\]

• Expand it: 

\[\sum_{k=1}^{n-1} (k^2 - k)(n-k) = \sum_{k=1}^{n-1} (k^2n - k^3 - kn + k^2)\]

• This is a polynomial expression, and for the limit as \( n \to \infty \), we focus on the leading terms: 

\[(n/3)(n^3) \approx \frac{n^3}{3}\]

1. Consider now the denominator: 

\[(1^3 + 2^3 + \ldots + n^3) - (1^2 + 2^2 + \ldots + n^2)\]

• Simplify using known formulas:
  • Cubes Sum: 

\[\sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2\]

• Squares Sum: 

\[\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\]

• Leading term of the denominator focusing on 

\[\left( \frac{n(n + 1)}{2} \right)^2 \approx \frac{n^4}{4}\]

1. For limits, compare the leading terms from numerator and denominator:

Numerator: 

\[\frac{n^3}{3}\]

Denominator: 

\[\frac{n^4}{4}\]

1. Now, calculate the limit:

\[\lim_{n \to \infty} \frac{\frac{n^3}{3}}{\frac{n^4}{4}} = \lim_{n \to \infty} \frac{4n^3}{3n^4} = \frac{4}{3n} \to \frac{1}{3}\]

 as \( n \to \infty \)

Hence, the correct answer is: 

\[\frac{1}{3}\]

.

Answer 2

\[ \lim_{n \to \infty} \sum_{r=1}^{n-1} (r^2 - r)(n - r) \]

\[ = \lim_{n \to \infty} \left( \sum_{r=1}^n r^3 - \sum_{r=1}^n r^2 \right) \]

\[ = \lim_{n \to \infty} \sum_{r=1}^{n-1} \left( -r^3 + r^2(n + 1) - nr \right) \]

\[ \lim_{n \to \infty} \left( \frac{n(n + 1)^2}{2} - \frac{n(n + 1)(2n + 1)}{6} - \frac{n^2(n - 1)}{2} \right) \]

Simplify further:

\[ \lim_{n \to \infty} \left( \frac{(n - 1)n}{2} + \frac{(n + 1)(n - 1)n(2n - 1) - n^2(n - 1)}{6} \right) \]

\[ \lim_{n \to \infty} \left[ \frac{n(n + 1)}{2} + \frac{n(n + 1)}{2} + \frac{2n + 1}{3} \right] \]

\[ \lim_{n \to \infty} \frac{n(n - 1)}{2} \left( -n(n - 1) + (n + 1)(2n - 1) \right) \]

\[ = \lim_{n \to \infty} \frac{n(n + 1)(3n^2 + 3n - 4n - 2)}{6} \]

\[ = \lim_{n \to \infty} \frac{(n - 1)(-3n^2 + 3 + 2(2n^2 + n - 1) - 6)}{(n + 1)(3n^2 - n - 2)} \]

\[ = \lim_{n \to \infty} \frac{(n - 1)(n^2 + 5n - 8)}{(n + 1)(3n^2 - n - 2)} = \frac{1}{3} \]