Animated Solution for Mathematics - Sequence and Series: $1^3-2^3+3^3-4^3+\cdots +9^3=$

Beyond Brute Force

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a tedious arithmetic nightmare. You see a series: $1^3-2^3+3^3-4^3+\cdots +9^3$.

Your instinct might be to start cubing numbers: $1,8,27,64,125,216,343,512,729$. You might start adding and subtracting them, hoping you don't make a sign error. Stop. Take a breath.

In the JEE Advanced arena, we don't calculate; we strategize. We look for the hidden structure.

The "What If" Strategy

Imagine for a moment that the world was perfect and all these terms were positive. The series would be a simple sum of cubes: $1^3+2^3+3^3+\cdots +9^3$. This is a standard, well-behaved mathematical object.

But our series is alternating. It has these pesky negative signs on the even terms: $-2^3,-4^3,-6^3,-8^3$.

Here is the mental pivot: If we take the sum of all positive cubes and subtract the even terms, we are left with the odd terms. But we need to subtract the even terms twice to account for the fact that they were originally negative.

If we define our sum as $S$, we can write:

Why the factor of $2$? Because in the first bracket, we have $2^3$ as a positive term. But in our original series, it needs to be $-2^3$.

To get from $+2^3$ to $-2^3$, we must subtract $2\times 2^3$. This logic holds for all even terms. We have successfully transformed a chaotic alternating series into two clean, manageable sums.

The Power of the Formula

Now, we invoke our most trusted weapon: the sum of cubes formula. You should have this etched into your memory:

This formula is elegant. It tells us that the sum of the first $n$ cubes is simply the square of the sum of the first $n$ natural numbers.

For our first group, where $n=9$, we calculate:

Calculating $45^2$ is a breeze if you know the trick for squaring numbers ending in $5$: multiply the first digit by the next integer ($4\times 5=20$) and append $25$. Thus, $45^2=2025$. We have our first piece of the puzzle.

Taming the Even Terms

Now, let's look at the second part: $2(2^3+4^3+6^3+8^3)$. This looks intimidating, but let's factor out the $2^3$ from each term. Recall that $2^3=8$.

So, we can write:

Substituting this back into our expression, we get:

We are back to the sum of cubes formula, but this time for $n=4$. Applying the formula again:

So, the total value of the even terms we need to subtract is $16\times 100=1600$.

The Final Convergence

We have arrived at the finish line. We have the total sum of all cubes ($2025$) and the value of the over-counted even terms ($1600$). The final step is a simple subtraction:

Look at that. We didn't need to cube $9$ or $8$ or $7$. We didn't need to perform a long, error-prone addition.

We used the structure of the series, the power of the summation formula, and the elegance of algebraic manipulation to arrive at 425. This is how you conquer JEE Advanced problems—not by brute force, but by seeing the patterns that others miss. Keep this mindset, and you will be unstoppable.