Animated Solution for Mathematics - Definite Integration: The value of ${\lim }_{n\to \infty }\frac{1}{n}{\sum }_{j=1}^n\frac{(2j-1)+8n}{(2j-1)+4n}$ is equal to :

Imagine you are standing on the edge of a vast, complex summation. You see a sequence of terms, each one slightly different, marching toward infinity. It looks like a chaotic, discrete mess.

But in the world of JEE Advanced, we have a superpower: Calculus. We can take this discrete staircase of a sum and turn it into a smooth, continuous ramp of an integral. This is the beauty of the Limit of a Sum.

Our problem is:

At first glance, it looks intimidating. But let's break it down. The key to unlocking this is to force the expression into the form $\frac{1}{n}\sum f(\frac{j}{n})$.

To get there, we need to create terms that look like $\frac{j}{n}$. Let's divide both the numerator and the denominator by $n$. This is our first step into the light:

Now, look at the term $\frac{2j-1}{n}$. We can rewrite this as $2(\frac{j}{n})-\frac{1}{n}$.

As $n$ marches toward infinity, that $\frac{1}{n}$ term shrinks to nothingness. It vanishes. So, we are left with a clean, elegant function of $x=\frac{j}{n}$.

Now, we apply the dictionary of calculus. The sum ${\sum }_{j=1}^n$ becomes the integral ${\int }_0^1$, the term $\frac{1}{n}$ becomes $dx$, and $\frac{j}{n}$ becomes $x$.

Our complex sum has transformed into a beautiful, simple definite integral:

Visualize this. Those infinite, tiny rectangles from our summation have now merged into a smooth, continuous area under the curve $f(x)=\frac{2x+8}{2x+4}$. We are now just finding the area under this curve from $x=0$ to $x=1$.

To integrate this, we don't need anything fancy. A simple algebraic trick will do. We want to simplify the numerator to match the denominator.

Let's rewrite $2x+8$ as $(2x+4)+4$. This allows us to split the fraction:

Now, we integrate term by term. The integral of $1$ is just $x$. For the second term, the integral of $\frac{4}{2x+4}$ is $2\ln (2x+4)$.

Evaluating this from $0$ to $1$ gives us:

Plugging in the limits:

Using the property $\ln A-\ln B=\ln (\frac{A}{B})$, we get:

And there it is. The final answer is $1+2\ln (\frac{3}{2})$. It is elegant, it is precise, and it is the result of turning a chaotic sum into a beautiful, continuous area.