Answer :
To solve the problem, we need to find the value of [tex]\(\lim _{n \rightarrow \infty} S_n\)[/tex] where [tex]\(S_n\)[/tex] is given by:
[tex]\[S_n = \sum_{k=1}^n \left[ \frac{24}{n^3} k^2 + \frac{12}{n^2} k + \frac{15}{n} \right]\][/tex]
Let's analyze the sum term-by-term.
### Step-by-Step Breakdown:
1. Separate the Sum:
[tex]\[S_n = \sum_{k=1}^n \frac{24}{n^3} k^2 + \sum_{k=1}^n \frac{12}{n^2} k + \sum_{k=1}^n \frac{15}{n}\][/tex]
2. Simplify Each Sum Separately:
#### First Term: [tex]\(\sum_{k=1}^n \frac{24}{n^3} k^2\)[/tex]
[tex]\[\sum_{k=1}^n \frac{24}{n^3} k^2 = \frac{24}{n^3} \sum_{k=1}^n k^2\][/tex]
The formula for the sum of squares of the first [tex]\(n\)[/tex] natural numbers is:
[tex]\[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\][/tex]
Substituting this in:
[tex]\[\frac{24}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{24}{6} \cdot \frac{(n+1)(2n+1)}{n^2} = 4 \cdot \frac{(n+1)(2n+1)}{n^2}\][/tex]
Simplifying further:
[tex]\[\frac{4(n+1)(2n+1)}{n^2} = 4 \cdot \left(\frac{(2n^2 + 3n + 1)}{n^2}\right) = 4 \cdot \left(2 + \frac{3}{n} + \frac{1}{n^2}\right)\][/tex]
As [tex]\(n \rightarrow \infty\)[/tex], [tex]\(\frac{3}{n} \rightarrow 0\)[/tex] and [tex]\(\frac{1}{n^2} \rightarrow 0\)[/tex]:
[tex]\[= 4 \cdot 2 = 8\][/tex]
#### Second Term: [tex]\(\sum_{k=1}^n \frac{12}{n^2} k\)[/tex]
[tex]\[\sum_{k=1}^n \frac{12}{n^2} k = \frac{12}{n^2} \sum_{k=1}^n k\][/tex]
The formula for the sum of the first [tex]\(n\)[/tex] natural numbers is:
[tex]\[\sum_{k=1}^n k = \frac{n(n+1)}{2}\][/tex]
Substituting this in:
[tex]\[\frac{12}{n^2} \cdot \frac{n(n+1)}{2} = \frac{12}{2} \cdot \frac{(n+1)}{n} = 6 \cdot \left(\frac{n+1}{n}\right)\][/tex]
Simplifying further:
[tex]\[6 \cdot \left(1 + \frac{1}{n}\right)\][/tex]
As [tex]\(n \rightarrow \infty\)[/tex], [tex]\(\frac{1}{n} \rightarrow 0\)[/tex]:
[tex]\[= 6 \cdot 1 = 6\][/tex]
#### Third Term: [tex]\(\sum_{k=1}^n \frac{15}{n}\)[/tex]
[tex]\[\sum_{k=1}^n \frac{15}{n} = \frac{15}{n} \sum_{k=1}^n 1 = \frac{15}{n} \cdot n = 15\][/tex]
3. Combine the Results:
Adding up the simplified limits, we get:
[tex]\[ \lim_{n \rightarrow \infty} S_n = 8 + 6 + 15 = 29 \][/tex]
Therefore, the value of [tex]\(\lim _{n \rightarrow \infty} S_n\)[/tex] is:
[tex]\[ \boxed{29} \][/tex]
[tex]\[S_n = \sum_{k=1}^n \left[ \frac{24}{n^3} k^2 + \frac{12}{n^2} k + \frac{15}{n} \right]\][/tex]
Let's analyze the sum term-by-term.
### Step-by-Step Breakdown:
1. Separate the Sum:
[tex]\[S_n = \sum_{k=1}^n \frac{24}{n^3} k^2 + \sum_{k=1}^n \frac{12}{n^2} k + \sum_{k=1}^n \frac{15}{n}\][/tex]
2. Simplify Each Sum Separately:
#### First Term: [tex]\(\sum_{k=1}^n \frac{24}{n^3} k^2\)[/tex]
[tex]\[\sum_{k=1}^n \frac{24}{n^3} k^2 = \frac{24}{n^3} \sum_{k=1}^n k^2\][/tex]
The formula for the sum of squares of the first [tex]\(n\)[/tex] natural numbers is:
[tex]\[\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\][/tex]
Substituting this in:
[tex]\[\frac{24}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{24}{6} \cdot \frac{(n+1)(2n+1)}{n^2} = 4 \cdot \frac{(n+1)(2n+1)}{n^2}\][/tex]
Simplifying further:
[tex]\[\frac{4(n+1)(2n+1)}{n^2} = 4 \cdot \left(\frac{(2n^2 + 3n + 1)}{n^2}\right) = 4 \cdot \left(2 + \frac{3}{n} + \frac{1}{n^2}\right)\][/tex]
As [tex]\(n \rightarrow \infty\)[/tex], [tex]\(\frac{3}{n} \rightarrow 0\)[/tex] and [tex]\(\frac{1}{n^2} \rightarrow 0\)[/tex]:
[tex]\[= 4 \cdot 2 = 8\][/tex]
#### Second Term: [tex]\(\sum_{k=1}^n \frac{12}{n^2} k\)[/tex]
[tex]\[\sum_{k=1}^n \frac{12}{n^2} k = \frac{12}{n^2} \sum_{k=1}^n k\][/tex]
The formula for the sum of the first [tex]\(n\)[/tex] natural numbers is:
[tex]\[\sum_{k=1}^n k = \frac{n(n+1)}{2}\][/tex]
Substituting this in:
[tex]\[\frac{12}{n^2} \cdot \frac{n(n+1)}{2} = \frac{12}{2} \cdot \frac{(n+1)}{n} = 6 \cdot \left(\frac{n+1}{n}\right)\][/tex]
Simplifying further:
[tex]\[6 \cdot \left(1 + \frac{1}{n}\right)\][/tex]
As [tex]\(n \rightarrow \infty\)[/tex], [tex]\(\frac{1}{n} \rightarrow 0\)[/tex]:
[tex]\[= 6 \cdot 1 = 6\][/tex]
#### Third Term: [tex]\(\sum_{k=1}^n \frac{15}{n}\)[/tex]
[tex]\[\sum_{k=1}^n \frac{15}{n} = \frac{15}{n} \sum_{k=1}^n 1 = \frac{15}{n} \cdot n = 15\][/tex]
3. Combine the Results:
Adding up the simplified limits, we get:
[tex]\[ \lim_{n \rightarrow \infty} S_n = 8 + 6 + 15 = 29 \][/tex]
Therefore, the value of [tex]\(\lim _{n \rightarrow \infty} S_n\)[/tex] is:
[tex]\[ \boxed{29} \][/tex]
