Question

lim n→∞ \(\frac{1}{n^3}\) \(\sum_{k=1}^{n} k^{2} =\)

 

• A. x
• B. \(\frac{x}{2}\)
• C. \(\frac{x}{3}\)
• D. \(\frac{x}{4}\)

Answer 1

The Correct Option is C

To solve the problem, we need to evaluate the limit $\lim_{n\to \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$.

1. Simplify the Expression:
Since $x$ is a constant, factor it out of the sum:
$\lim_{n\to \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x) = \lim_{n\to \infty} \frac{x}{n^3} \sum_{k=1}^n k^2$.

2. Apply the Sum of Squares Formula:
We know $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.
Substitute this into the limit:
$\lim_{n\to \infty} \frac{x}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \lim_{n\to \infty} \frac{x}{6n^3} \cdot n(2n^2 + 3n + 1)$.

3. Simplify the Limit:
Rewrite the expression:
$\lim_{n\to \infty} \frac{x(2n^3 + 3n^2 + n)}{6n^3} = \lim_{n\to \infty} \frac{x}{6} \left( 2 + \frac{3}{n} + \frac{1}{n^2} \right)$.
As $n \to \infty$, $\frac{3}{n} \to 0$ and $\frac{1}{n^2} \to 0$, so the limit becomes:
$\frac{x}{6} \cdot 2 = \frac{x}{3}$.

Final Answer:
The limit is $\frac{x}{3}$.