Animated Solution for Mathematics - Binomial Theorem: If the sum of the coefficients of all even powers of $x$ in the product $(1+x+x^2+\cdots +x^{2n})(1-x+x^2-\cdots +x^{2n})$ is 61, then $n$ is equal to

Imagine you are sitting in the examination hall. You see a problem involving the product of two long polynomials: $(1+x+x^2+\cdots +x^{2n})$ and $(1-x+x^2-\cdots +x^{2n})$.

Your first instinct might be to multiply them out. Stop! That is the trap.

If you try to expand this, you will be lost in a sea of terms, and you will likely run out of time before you even reach the first coefficient. In JEE Advanced, the most complex-looking problems often have the most elegant, hidden solutions. Today, we are going to learn how to master the polynomial without ever expanding it.

Let us define our product as a single polynomial, $P(x)$. We can write it as:

Our objective is to find the sum of the coefficients of the even powers of $x$. That is, we want to calculate $S_{even}=a_0+a_2+a_4+\dots$.

We do not need to know what $a_0$ or $a_2$ are individually. We only need their sum. This is a shift in perspective—we are looking for the aggregate behavior of the polynomial, not the individual components.

How do we isolate the even terms? We use the power of symmetry. Consider what happens when we evaluate $P(1)$:

This gives us the sum of all coefficients. Now, consider $P(-1)$:

Look at that! The odd-indexed coefficients have become negative, while the even-indexed coefficients remain positive. If we add these two equations together, the odd terms cancel out perfectly, leaving us with twice the sum of the even terms:

Thus, our master formula is born:

Now, let us evaluate $P(1)$ and $P(-1)$ using the original expression. First, $P(1)$:

The first bracket contains $2n+1$ terms, all equal to $1$, so it sums to $2n+1$. The second bracket is an alternating series of $1$ and $-1$. Since there are $2n+1$ terms (an odd number), the pairs cancel out, leaving just $1$.

Thus, $P(1)=(2n+1)\times 1=2n+1$.

Next, $P(-1)$:

In the first bracket, the terms become $1-1+1-1\dots$, which again sums to $1$. In the second bracket, the terms become $1+1+1+\cdots +1$. Since there are $2n+1$ terms, this sums to $2n+1$.

Thus, $P(-1)=1\times (2n+1)=2n+1$.

We have our values. Substituting them into our master formula:

The problem states that this sum is $61$. Therefore:

We have arrived at the solution. Notice how we never expanded the polynomial? We used the properties of the function itself to bypass the brute force.

This is the mindset of a JEE topper: look for the symmetry, use the identity, and let the algebra do the heavy lifting for you. Keep this trick in your arsenal, and you will be ready for any polynomial challenge that comes your way!