Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{n\to \infty }\frac{(1^2-1)(n-1)+(2^2-2)(n-2)+\cdots +((n-1{)}^2-(n-1))\cdot 1}{(1^3+2^3+\cdots +n^3)-(1^2+2^2+\cdots +n^2)}$ is equal to :

A Journey Through Limits and Summations

Welcome, future engineer. Today, we stand before a problem that, at first glance, looks like a chaotic mess of algebra. You see a massive fraction, a series of products in the numerator, and a difference of power sums in the denominator.

It is designed to intimidate. It is designed to make you reach for a calculator or, worse, give up. But I want you to take a deep breath.

In the world of JEE Advanced, we do not fear complexity; we dismantle it. We are going to break this beast down, piece by piece, until it reveals its elegant, simple core.

The Anatomy of the Numerator

Let us look at the numerator: $(1^2-1)(n-1)+(2^2-2)(n-2)+\cdots +((n-1{)}^2-(n-1))\cdot 1$.

Do you see the rhythm? Mathematics is, at its heart, the study of patterns. The first term is $(1^2-1)(n-1)$ and the second is $(2^2-2)(n-2)$.

If we let $r$ represent the index of the term, the general term $T_r$ is clearly $(r^2-r)(n-r)$.

Now, where does this dance begin and end? The first term corresponds to $r=1$. The last term, $((n-1{)}^2-(n-1))\cdot 1$, corresponds to $r=n-1$.

So, our numerator is simply the summation:

Before we rush to calculate, let us expand this. It is a polynomial in $r$:

Now, we distribute the summation operator. Remember, $n$ is a constant relative to $r$:

This is the heart of the problem. We have transformed a terrifying string of numbers into a structured, solvable algebraic expression.

The Power of Standard Formulas

We know our trusty tools: the standard summation formulas.

1. ${\sum }_{r=1}^{k}r=\frac{k(k+1)}{2}$ 2. ${\sum }_{r=1}^k{r}^2=\frac{k(k+1)(2k+1)}{6}$ 3. ${\sum }_{r=1}^k{r}^3={\left(\frac{k(k+1)}{2}\right)}^2$

Here is where many students stumble. They blindly plug in $n$. But our summation goes to $n-1$. So, we must substitute $k=n-1$.

For the numerator, we are looking for the behavior as $n\to \infty$. We do not need to expand every bracket. We only need the coefficient of the highest power of $n$.

Let us look at the terms:

- The term $-\sum r^3$ will behave like $-\frac{n^4}{4}$. - The term $(n+1)\sum r^2$ will behave like $n\cdot \frac{2n^3}{6}=\frac{n^4}{3}$. - The term $-n\sum r$ will behave like $-n\cdot \frac{n^2}{2}=-\frac{n^3}{2}$.

Combining these, the leading coefficient of $n^4$ in the numerator is $-\frac{1}{4}+\frac{1}{3}=\frac{1}{12}$.

The Denominator and the Final Victory

Now, the denominator $D=(1^3+2^3+\cdots +n^3)-(1^2+2^2+\cdots +n^2)$. This is much friendlier. It is simply ${\sum }_{r=1}^n{r}^3-{\sum }_{r=1}^n{r}^2$.

Using our formulas with $k=n$, the leading term of $\sum r^3$ is $\frac{n^4}{4}$. The leading term of $\sum r^2$ is $\frac{n^3}{3}$.

As $n\to \infty$, the $n^4$ term dominates. Thus, the coefficient of $n^4$ in the denominator is simply $\frac{1}{4}$.

We are at the finish line. The limit of the ratio of two polynomials as $n\to \infty$ is simply the ratio of the coefficients of their highest powers:

Look at that. The chaos has vanished. The complexity has collapsed into a single, beautiful fraction: $\frac{1}{3}$.

You didn't just solve a problem; you navigated a labyrinth of algebra and emerged victorious. Keep this mindset—break the big into the small, identify the pattern, and trust the math. You are ready for the next challenge.