Question

\(\lim\limits_{n \rightarrow \infin}\frac{6}{n+2}\left\{(2+\frac{1}{n})^2+(2+\frac{2}{n})^2+...+(2+\frac{n-1}{2})^2\right\}\) equals

• A. 38
• B. 36
• C. 32
• D. 30

Answer 1

The Correct Option is A

To solve the given problem, we need to evaluate the following limit:

\[\lim\limits_{n \rightarrow \infty}\frac{6}{n+2}\left\{(2+\frac{1}{n})^2+(2+\frac{2}{n})^2+\ldots+(2+\frac{n-1}{n})^2\right\}\]

First, let's rewrite the expression inside the brackets. It is a sum of squares:

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

Each term in the expansion can be expressed as:

\[\left(2+\frac{k}{n}\right)^2 = 4 + \frac{4k}{n} + \frac{k^2}{n^2}\]

Thus, the sum becomes:

\[\sum_{k=1}^{n-1} \left(4 + \frac{4k}{n} + \frac{k^2}{n^2}\right)\]

Separating the sums, we have:

\[4(n-1) + \frac{4}{n} \sum_{k=1}^{n-1} k + \frac{1}{n^2} \sum_{k=1}^{n-1} k^2\]

Using the formulas for the sums, \(\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}\) and \(\sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}\), we substitute to get:

\[4(n-1) + \frac{4}{n} \cdot \frac{(n-1)n}{2} + \frac{1}{n^2} \cdot \frac{(n-1)n(2n-1)}{6}\]

Simplifying each part:

• First term: \(4(n-1) = 4n - 4\)
• Second term: \(\frac{4(n-1)n}{2n} = 2(n-1)\)
• Third term: \(\frac{(n-1)(2n-1)}{6n} \approx \frac{2n^2 - 3n + 1}{6n} = \frac{n}{3} - \frac{1}{2} + \frac{1}{6n}\)

Plug these into the original expression:

\[\frac{6}{n+2} \left(4n - 4 + 2(n-1) + \left(\frac{n}{3} - \frac{1}{2}\right)\right)\]

Simplify the expression further:

• Combine similar terms: \(6n + \frac{n}{3} - 6\)
• The limit becomes: \(im_{n \to \infty} \frac{38n}{n+2} - \frac{1}{n+2}\). When calculated, this approaches 38 as \(n \to \infty\)

Therefore, the limit is 38.