Animated Solution for Mathematics - Complex Numbers: The value of the expression $1\cdot (2-\omega )(2-{\omega }^2)+2\cdot (3-\omega )(3-{\omega }^2)+\cdots +(n-1)\cdot (n-\omega )(n-{\omega }^2)$, where $\omega$ is an imaginary cube root of unity, is $\dots$

Unmasking the Series

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a complex nightmare.

You see terms involving $\omega$, the imaginary cube root of unity, and your instinct might be to panic. But I want you to take a deep breath.

In the world of JEE Advanced, problems are often wolves in sheep's clothing. They look intimidating, but once you strip away the disguise, they reveal a beautiful, simple structure. Let's embark on this journey together.

Pattern Recognition

Look at the series: $1\cdot (2-\omega )(2-{\omega }^2)+2\cdot (3-\omega )(3-{\omega }^2)+\cdots +(n-1)\cdot (n-\omega )(n-{\omega }^2)$.

The first thing we must do is identify the general term, $T_r$. Notice the pattern: when the first factor is $1$, the inner terms involve $2$. When the first factor is $2$, the inner terms involve $3$.

It is clear that for the $r$-th term, the first factor is $r$, and the inner terms are $(r+1-\omega )$ and $(r+1-{\omega }^2)$. So, our general term is:

The Magic Wand Identity

Here is where the magic happens. Whenever you see $(x-\omega )(x-{\omega }^2)$, you should immediately think of the cube roots of unity.

We know that $x^3-1=(x-1)(x-\omega )(x-{\omega }^2)$. By dividing by $(x-1)$, we get the identity:

This is our magic wand. It allows us to banish the complex numbers entirely. By substituting $x=r+1$, our general term becomes:

The Algebraic Cleanup

Now, don't let the algebra scare you. It is just a matter of careful expansion.

Expanding $(r+1{)}^2$ gives $r^2+2r+1$. Adding the remaining terms, we get:

This simplifies to $T_r=r(r^2+3r+3)$. Distributing the $r$, we arrive at the clean, beautiful polynomial:

See? The complex numbers are gone, and we are left with a standard polynomial that we can sum with ease.

The Summation Trap

We need to find the sum $S={\sum }_{r=1}^{n-1}{T}_r$. Here is the most important lesson of the day: check your limits!

The series ends at $(n-1)$, so our summation must go from $r=1$ to $n-1$. If you sum to $n$, you will lose marks.

Using the linearity of summation, we split this into:

We then apply the standard formulas, remembering to replace $n$ with $n-1$ in each one:

The Final Victory

Finally, we simplify. By factoring out $\frac{n(n-1)}{4}$, the expression inside the bracket becomes:

Expanding this, we get $n^2-n+4n-2+6$, which simplifies to $n^2+3n+4$. Thus, our final answer is:

We have conquered the beast! Remember, the key was not brute force, but recognizing the underlying structure. Keep practicing, and you will start to see these patterns everywhere. You have got this!