Then, the value of lim f (n) is equal to : 17100 A 3 + 4/3 * ln(7) 4 - 3/4 * ln(7/3) 4 - 4/3 * ln(7/3) 3 + 3/4 * ln(7) f(n)= n + (16 + 5n - 3n ^ 2)/(4n + 3n ^ 2) + (32 + n - 3n ^ 2)/(8n + 3n ^ 2) + (48 - 3n - 3n ^ 2)/(12n + 3n ^ 2) +***+ 25n-7n^ 2 7n^ 2 .

1. Recognising the pattern

Write the function more neatly for a fixed positive integer $n$:

Why does that numerator look so weird? Because the popular trick “add 1 and subtract 1” is hidden inside it!

2. Kill the big–$n$ explosion

Each term inside the sum is roughly $-1$ (because the $-3n^{2}$ and $3n^{2}$ dominate). Summing $-1$ over $n$ terms gives $-n$, which cancels the lonely leading $+n$. So everything infinite disappears and you are left with a finite expression:

Therefore

3. Spotting the Riemann sum

Factor out $n^{2}$ from numerator and denominator:

Put $x_k = \dfrac{k}{n}$ (right-end points of $n$ equal sub-intervals on $[0,1]$). Then

Hence

This is exactly the right-Riemann sum for

4. Compute the integral

Do a quick long division:

So

5. Limit value

Therefore

Among the four options, that matches Option B.

What is the question?

You have a long-looking fraction–sum that depends on a number $n$. The question asks, “When $n$ becomes a really, really huge number, what does the whole expression settle down to?”

How to think about it (like a story)

1. Picture $n$ as the number of tiny steps along a road from 0 to 1.
2. Each fraction in the sum is like measuring the height of a curve at one of those steps.
3. Adding all those little heights and multiplying by the step–width ($1/n$) is the same trick we use to turn a sum into an area under a curve (this is what grown-ups call a Riemann sum).
4. When $n$ is gigantic, the road has millions of tiny steps, so the sum is almost exactly the integral of that curve from 0 to 1.
5. Once you spot the right curve, doing the integral is easy and gives a nice clean number with a small natural‐log piece.

That final number is the answer!

Step–by–step calculation

1. Rewrite $f(n)$

   $$f(n)=n+\sum_{k=1}^{n}\frac{16k+(9-4k)n-3n^2}{4kn+3n^2}.$$

2. Separate out the “$+1$”

   $$\begin{aligned} T_k &= \frac{16k+(9-4k)n-3n^2}{4kn+3n^2},\\ T_k+1 &= \frac{16k+9n}{4kn+3n^2}. \end{aligned}$$

   Hence

   $$f(n)=n-n+\sum_{k=1}^{n}\frac{16k+9n}{4kn+3n^2}=\sum_{k=1}^{n}\frac{16k+9n}{4kn+3n^2}.$$

3. Normalise each term
   Divide top and bottom by $n^{2}$ and set $x_k=\dfrac{k}{n}$:

   $$f(n)=\sum_{k=1}^{n}\frac{1}{n}\,\frac{16x_k+9}{4x_k+3}.$$

4. Recognise the Riemann sum
   As $n\to\infty$ this becomes the integral

   $$\int_{0}^{1}\frac{16x+9}{4x+3}\,dx.$$

5. Integrate
   Long-divide:

   $$\frac{16x+9}{4x+3}=4-\frac{3}{4x+3}.$$

   Then

   $$\begin{aligned} \int_{0}^{1}\left(4-\frac{3}{4x+3}\right)dx&=4\left[x\right]_{0}^{1}-\frac{3}{4}\left[\ln(4x+3)\right]_{0}^{1}\\[6pt] &=4-\frac{3}{4}\ln\left(\frac{7}{3}\right). \end{aligned}$$

6. Final answer

   $$\boxed{\lim_{n\to\infty}f(n)=4-\frac{3}{4}\ln\!\left(\frac{7}{3}\right)}.$$

   This is Option (2) (or B) from the list.