The user wants me to solve the limit problem: For a positive integer $n$, define $f(n)=n+\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+\frac{48-3 n-3 n^{2}}{12 n+3 n^{2}}+\cdots+\frac{25 n-7 n^{2}}{7 n^{2}} .$ Then, find the value of $\mathop {\lim }\limits_{n \to \infty } f\left( n \right)$.

Step 1: Identify the general term of the summation. Let the sum be denoted by $S_n$. Then $f(n) = n + S_n$. The sum is of the form $\sum_{r=1}^{k} T_r$. Let's analyze the $r$-th term, $T_r$.

The denominators are:

• $r=1: 4n+3n^2 = 4(1)n+3n^2$
• $r=2: 8n+3n^2 = 4(2)n+3n^2$
• $r=3: 12n+3n^2 = 4(3)n+3n^2$ The general form of the denominator is $D_r = 4rn + 3n^2$.

The numerators are:

• $r=1: 16+5n-3n^2$
• $r=2: 32+n-3n^2$
• $r=3: 48-3n-3n^2$ Let's find the pattern for the numerator, $N_r$. The terms independent of $n$ are $16, 32, 48, \ldots$, which form an arithmetic progression. The $r$-th term is $16r$. The coefficients of $n$ are $5, 1, -3, \ldots$, which form an arithmetic progression with first term $5$ and common difference $-4$. The $r$-th term is $5 + (r-1)(-4) = 9-4r$. The coefficient of $n^2$ is constant at $-3$ for the first few terms. So, the general form of the numerator appears to be $N_r = 16r + (9-4r)n - 3n^2$.

Step 2: Determine the number of terms in the sum. Let the number of terms be $k$. The last term is given as $\frac{25 n-7 n^{2}}{7 n^{2}}$. Let's check if this term fits the general pattern for some value of $r=k$. Let's check the denominator first. If $D_k = 7n^2$, then $4kn + 3n^2 = 7n^2$. This gives $4kn = 4n^2$, which implies $k=n$. So there are $n$ terms in the sum. Now let's check if the numerator matches for $r=n$. $N_n = 16n + (9-4n)n - 3n^2 = 16n + 9n - 4n^2 - 3n^2 = 25n - 7n^2$. This matches the numerator of the last term. So the sum is from $r=1$ to $n$.

The sum is $S_n = \sum_{r=1}^{n} \frac{16r + (9-4r)n - 3n^2}{4rn + 3n^2}$.

Step 3: Simplify the expression for $f(n)$. Let's simplify the general term $T_r$: $T_r = \frac{16r + 9n - 4rn - 3n^2}{4rn + 3n^2}$ We can rewrite the numerator to relate it to the denominator: $N_r = -(4rn + 3n^2) + 16r + 9n$ So, $T_r = \frac{-(4rn + 3n^2) + 16r + 9n}{4rn + 3n^2} = -1 + \frac{16r + 9n}{4rn + 3n^2}$ Now, we can write $f(n)$ as: $f(n) = n + S_n = n + \sum_{r=1}^{n} \left(-1 + \frac{16r + 9n}{4rn + 3n^2}\right)$ $f(n) = n + \left(\sum_{r=1}^{n} -1\right) + \sum_{r=1}^{n} \frac{16r + 9n}{4rn + 3n^2}$ $f(n) = n - n + \sum_{r=1}^{n} \frac{16r + 9n}{4rn + 3n^2} = \sum_{r=1}^{n} \frac{16r + 9n}{4rn + 3n^2}$

Step 4: Evaluate the limit using Riemann sums. The limit is $\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{16r + 9n}{4rn + 3n^2}$. To convert this to a definite integral, we need to express the term in the form $\frac{1}{n} F(\frac{r}{n})$. Divide the numerator and denominator of the general term by $n^2$: $\frac{16r + 9n}{4rn + 3n^2} = \frac{\frac{16r}{n^2} + \frac{9n}{n^2}}{\frac{4rn}{n^2} + \frac{3n^2}{n^2}} = \frac{\frac{16}{n}\frac{r}{n} + \frac{9}{n}}{4\frac{r}{n} + 3} = \frac{1}{n} \frac{16\frac{r}{n} + 9}{4\frac{r}{n} + 3}$ So, we have the limit of a Riemann sum: $\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \frac{16(r/n) + 9}{4(r/n) + 3}$ This limit is equal to the definite integral $\int_{0}^{1} F(x) dx$, where $x$ replaces $r/n$. Here, $F(x) = \frac{16x+9}{4x+3}$.

Step 5: Calculate the definite integral. $L = \int_{0}^{1} \frac{16x+9}{4x+3} dx$ We can simplify the integrand by polynomial division: $16x+9 = 4(4x+3) - 12 + 9 = 4(4x+3) - 3$ So, $\frac{16x+9}{4x+3} = \frac{4(4x+3) - 3}{4x+3} = 4 - \frac{3}{4x+3}$. Now, we integrate: $L = \int_{0}^{1} \left(4 - \frac{3}{4x+3}\right) dx$ $L = \left[ 4x - 3 \cdot \frac{1}{4} \ln|4x+3| \right]_0^1 = \left[ 4x - \frac{3}{4} \ln(4x+3) \right]_0^1$ Evaluate at the limits: $L = \left(4(1) - \frac{3}{4}\ln(4(1)+3)\right) - \left(4(0) - \frac{3}{4}\ln(4(0)+3)\right)$ $L = \left(4 - \frac{3}{4}\ln(7)\right) - \left(0 - \frac{3}{4}\ln(3)\right)$ $L = 4 - \frac{3}{4}\ln(7) + \frac{3}{4}\ln(3)$ $L = 4 - \frac{3}{4}(\ln(7) - \ln(3))$ $L = 4 - \frac{3}{4}\ln\left(\frac{7}{3}\right)$

Comparing this result with the given options: A: $3+\frac{4}{3} \log _{e} 7$ B: $4-\frac{3}{4} \log _{e}\left(\frac{7}{3}\right)$ C: $4-\frac{4}{3} \log _{e}\left(\frac{7}{3}\right)$ D: $3+\frac{3}{4} \log _{e} 7$

Our result matches option B.