Animated Solution for Mathematics - Sequence and Series: For any odd integer $n\ge 1$, $n^3-(n-1{)}^3+\cdots +(-1{)}^{n-1}{1}^3=\dots$.

Welcome, future engineer. Today, we are going to dismantle a problem that looks like a chaotic mess of alternating signs but is actually a masterpiece of algebraic symmetry.

When you first look at the series $S=n^3-(n-1{)}^3+(n-2{)}^3-\cdots +(-1{)}^{n-1}{1}^3$, it is natural to feel a bit overwhelmed. We have alternating signs, cubes, and a variable $n$ that changes the very structure of the series.

But remember, in JEE Advanced, complexity is often just a veil for elegance. Let us pull back that veil together.

First, we must respect the constraints. The problem tells us $n$ is an odd integer. This is our anchor.

If $n$ is odd, then $n-1$ is even. Look at the last term: $(-1{)}^{n-1}{1}^3$. Since $n-1$ is even, $(-1{)}^{n-1}$ is positive.

This confirms that our series ends with a positive $1^3$. We are not dealing with a random sequence; we are dealing with a structured, alternating sum of cubes.

Now, how do we handle the alternating signs? We do not. Instead, we transform the problem.

We want to turn this alternating mess into a continuous, beautiful sum of cubes. We know the formula for the sum of the first $n$ cubes is:

To get there, we need all terms to be positive. So, we add the even terms $(n-1{)}^3,(n-3{)}^3,\dots ,2^3$ to our series.

But we cannot just add numbers to an equation without consequences. To keep the balance, we must subtract those same terms. Because they were originally negative in our alternating series, adding them once makes them zero, and subtracting them again makes them negative.

Thus, we are effectively subtracting them twice. Our series becomes:

We have split the problem into two manageable parts. The first part is the sum of all cubes from $1$ to $n$, which is simply:

Now, look at the second part: $2[(n-1{)}^3+(n-3{)}^3+\cdots +2^3]$. Every base here is even. We can factor out a $2^3$, which is $8$, from every term.

This transforms the bracket into:

Let $m=\frac{n-1}{2}$. The bracket is now just the sum of cubes from $1^3$ to $m^3$, which is $[\frac{m(m+1)}{2}{]}^2$. Substituting $m$ back, we get:

We are at the finish line. Combining our parts, we have:

Simplifying the second term, $16\cdot \frac{(n-1{)}^2(n+1{)}^2}{64}$ becomes $\frac{(n-1{)}^2(n+1{)}^2}{4}$. Now, both terms share a common factor of $\frac{(n+1{)}^2}{4}$.

Factoring this out, we are left with:

Using the difference of squares identity, $n^2-(n-1{)}^2$ becomes $(n-(n-1))(n+(n-1))$, which is $1\cdot (2n-1)$.

The final result is:

Look at that elegance! We took a terrifying alternating series and reduced it to a simple polynomial. This is the power of mathematical strategy. Keep practicing, keep questioning, and never let the complexity intimidate you.