Animated Solution for Mathematics - Binomial Theorem: Prove that \sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0, where $k=(3n)/2$ and $n$ is an even positive integer.

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving an identity; we are uncovering a hidden symmetry between algebra and geometry.

The problem asks us to prove that:

where $k=\frac{3n}{2}$ and $n$ is an even positive integer. At first glance, this summation looks like a daunting wall of binomial coefficients and powers. I want you to see it as a shadow cast by a complex number.

The Geometric Insight

To crack this, we need to construct a complex number $z$ that, when expanded, mirrors the structure of our summation. Look closely at the term $(-3{)}^{r-1}$. This is the result of squaring $i\sqrt{3}$.

If we define $z=1+i\sqrt{3}$, we are setting the stage for a beautiful transformation. Let us plot this on the Argand plane. The real part is $1$, and the imaginary part is $\sqrt{3}$.

The magnitude is $|z|=\sqrt{1^2+(\sqrt{3}{)}^2}=2$, and the argument is $\theta ={\tan }^{-1}(\frac{\sqrt{3}}{1})=\frac{\pi }{3}$. Thus, in polar form, we have:

The Power of De Moivre

Now, let us raise this complex number to the power of $3n$. This is where De Moivre's Theorem becomes our most powerful tool.

We have $z^{3n}=[2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}){]}^{3n}$. Applying the theorem, the magnitude becomes $2^{3n}$, and the angle is multiplied by $3n$:

Here is the moment of truth. The problem tells us that $n$ is an even positive integer. For any even integer $n$, $\sin (n\pi )=0$ and $\cos (n\pi )=1$.

Therefore, $z^{3n}=2^{3n}(1+i\cdot 0)=2^{3n}$. The imaginary part of $z^{3n}$ is exactly zero. This is our anchor.

The Binomial Bridge

Now, let us look at the same expression through the lens of the Binomial Theorem. We expand $(1+i\sqrt{3}{)}^{3n}$ as:

We are only interested in the imaginary part of this expansion. Notice that $(i\sqrt{3}{)}^j$ is real when $j$ is even and imaginary when $j$ is odd.

To capture the imaginary part, we substitute $j=2r-1$, where $r$ ranges from $1$ to $k=\frac{3n}{2}$. The imaginary part is:

The Grand Finale

Let us simplify the term $(i\sqrt{3}{)}^{2r-1}$. We can write it as $(i\sqrt{3}{)}^{2(r-1)}\cdot (i\sqrt{3})$. Since $(i\sqrt{3}{)}^2=-3$, this becomes $(-3{)}^{r-1}\cdot i\sqrt{3}$.

Substituting this back into our summation, we get:

We already established that $Im(z^{3n})=0$. Therefore:

Since $\sqrt{3} eq 0$, we can safely divide by it, leaving us with the beautiful result:

We have successfully bridged the gap between complex geometry and binomial algebra. Take a moment to appreciate the elegance of that cancellation—it is the hallmark of a problem well solved.