Animated Solution for Mathematics - Definite Integration: For positive integer $n$, define $f(n)=n+\frac{16+5n-3n^2}{4n+3n^2}+\frac{32+n-3n^2}{8n+3n^2}+\frac{48-3n-3n^2}{12n+3n^2}+\cdots +\frac{25n-7n^2}{7n^2}$. Then, the value of ${\lim }_{n\to \infty }f(n)$ is equal to

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of terms.

You see a series, a limit, and a function $f(n)$ that seems to grow in complexity with every step. But beneath this surface, there is a hidden order waiting to be revealed. Let us embark on this journey together.

Imagine you are standing before a long, winding path of fractions: $f(n)=n+\frac{16+5n-3n^2}{4n+3n^2}+\cdots +\frac{25n-7n^2}{7n^2}$. Your first instinct might be to panic. Don't.

Instead, look for the rhythm. In mathematics, as in music, patterns are everything. If we look at the denominators, we see $4n+3n^2,8n+3n^2,12n+3n^2$, and so on. They are all of the form $4rn+3n^2$.

Now, look at the numerators. The constant terms $16,32,48$ are clearly multiples of $16r$. The coefficients of $n$ follow their own arithmetic progression. By carefully extracting these, we define the general term of our series as:

This is where the magic happens. We have a fraction that looks intimidating, but notice the denominator $4rn+3n^2$. If we look at the numerator, we see $-3n^2$ and $-4rn$.

If we group these, we get $-(4rn+3n^2)$. This is a stroke of genius! We can rewrite the numerator as $16r+9n-(4rn+3n^2)$.

When we split the fraction, we get:

Now, look back at our function $f(n)$. We have $n+{\sum }_{r=1}^n(\cdots -1)$. Since the sum of $1$ from $r=1$ to $n$ is simply $n$, the leading $n$ and the $-n$ from the summation cancel out perfectly. We are left with a much cleaner expression:

We are now standing at the threshold of the Riemann Sum. To cross it, we need to transform our discrete sum into a continuous integral. We divide the numerator and denominator by $n^2$ to create terms of $\frac{r}{n}$:

As $n\to \infty$, the sum $\frac{1}{n}{\sum }_{r=1}^{n}g(\frac{r}{n})$ becomes the definite integral ${\int }_0^{1}g(x)dx$. Our problem has transformed from a tedious summation into a beautiful integral:

To solve ${\int }_0^1\frac{16x+9}{4x+3}dx$, we use the same trick we used earlier: force the numerator to look like the denominator. We write $16x+9$ as $4(4x+3)-3$.

The integral splits into two parts:

The first part is trivial: $4$. The second part is a standard logarithmic integral: $\frac{3}{4}\ln |4x+3|$. Evaluating this from $0$ to $1$ gives us:

Using the property $\ln a-\ln b=\ln (\frac{a}{b})$, we arrive at our final, elegant answer:

You see? What started as a daunting wall of numbers was merely a puzzle waiting for the right perspective. Keep this curiosity alive, and no problem will ever be too large to solve.