To solve the problem, we need to find the limit of the ratio of the sums of the terms of two sequences as \( n \) approaches infinity. Let's break down the solution step by step.
### Step 1: Identify the nth terms of the sequences
The first sequence is \( 3, 6, 11, 18, 27, \ldots \). We can observe the pattern in the differences:
- \( 6 - 3 = 3 \)
- \( 11 - 6 = 5 \)
- \( 18 - 11 = 7 \)
- \( 27 - 18 = 9 \)
The differences are \( 3, 5, 7, 9, \ldots \), which is an arithmetic sequence with a common difference of \( 2 \). Thus, the \( n^{th} \) term \( f(n) \) can be expressed as:
\[
f(n) = n^2 + 2
\]
### Step 2: Find the sum of the first n terms \( F(n) \)
The sum of the first \( n \) terms \( F(n) \) is given by:
\[
F(n) = \sum_{k=1}^{n} f(k) = \sum_{k=1}^{n} (k^2 + 2)
\]
Using the formula for the sum of squares and the sum of a constant:
\[
F(n) = \frac{n(n+1)(2n+1)}{6} + 2n
\]
### Step 3: Identify the nth terms of the second sequence
The second sequence is \( 3, 7, 13, 21, \ldots \). The differences are:
- \( 7 - 3 = 4 \)
- \( 13 - 7 = 6 \)
- \( 21 - 13 = 8 \)
The differences are \( 4, 6, 8, \ldots \), which is also an arithmetic sequence with a common difference of \( 2 \). Thus, the \( n^{th} \) term \( g(n) \) can be expressed as:
\[
g(n) = n^2 + n + 1
\]
### Step 4: Find the sum of the first n terms \( G(n) \)
The sum of the first \( n \) terms \( G(n) \) is given by:
\[
G(n) = \sum_{k=1}^{n} g(k) = \sum_{k=1}^{n} (k^2 + k + 1)
\]
Using the formulas for the sum of squares, the sum of the first \( n \) natural numbers, and the sum of a constant:
\[
G(n) = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n
\]
### Step 5: Calculate the limit
Now, we need to find:
\[
\lim_{n \to \infty} \frac{F(n)}{G(n)}
\]
Substituting the expressions for \( F(n) \) and \( G(n) \):
\[
\lim_{n \to \infty} \frac{\frac{n(n+1)(2n+1)}{6} + 2n}{\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n}
\]
As \( n \) approaches infinity, the dominant terms in both the numerator and denominator are \( \frac{n^3}{3} \) and \( \frac{n^3}{3} \) respectively. Therefore, we can simplify:
\[
\lim_{n \to \infty} \frac{F(n)}{G(n)} = \frac{1}{1} = 1
\]
### Final Answer
Thus, the limit is:
\[
\lim_{n \to \infty} \frac{F(n)}{G(n)} = 1
\]
