Animated Solution for Mathematics - Binomial Theorem: If the sum of the coefficients of all even powers of $x$ in the product $(1+x+x^2+...+x^{2n})(1-x+x^2-x^3+...+x^{2n})$ is 61, then $n$ is equal to . . . . .

Imagine you are standing before a massive, intimidating wall of algebra. You see the product:

Your first instinct might be to multiply it out, to distribute every term, and to create a sprawling mess of exponents. Stop. In the world of JEE Advanced, brute force is rarely the path to victory; the path to victory is insight.

We define our polynomial as $P(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots +a_{4n}{x}^{4n}$. The question asks for the sum of the coefficients of all even powers of $x$.

We define our target sum as $S=a_0+a_2+a_4+\cdots +a_{4n}$. If you try to expand this, you will lose hours. Instead, we look for symmetry.

There is a beautiful trick in polynomial theory. If we want to isolate even coefficients, we use the parity of the variable $x$.

When we evaluate $P(1)$, every $x$ becomes $1$, so:

This represents the sum of all coefficients. Now, consider $P(-1)$. Every even power $x^{2k}$ remains positive, but every odd power $x^{2k+1}$ becomes negative. Thus:

When we add these two, $P(1)+P(-1)$, the odd terms cancel out perfectly, leaving us with $2(a_0+a_2+a_4+\dots )$. Therefore, our target sum is:

Let us evaluate $P(1)$. The first bracket is $(1+1+1^2+\cdots +1^{2n})$. Since there are $2n+1$ terms, this sum is $2n+1$.

The second bracket is $(1-1+1-1+\cdots +1)$. The pairs cancel out, leaving just the last term, which is $1$. So, $P(1)=(2n+1)\times 1=2n+1$.

Now, for $P(-1)$. The first bracket becomes $(1-1+1-1+\cdots +1)$, which equals $1$. The second bracket becomes $(1-(-1)+(-1{)}^2-(-1{)}^3+\cdots +(-1{)}^{2n})$, which simplifies to $(1+1+1+\cdots +1)$.

Again, with $2n+1$ terms, this sum is $2n+1$. So, $P(-1)=1\times (2n+1)=2n+1$.

We have our values. Substituting them into our formula for $S$:

The problem states that this sum is $61$. Therefore, we set $2n+1=61$.

Subtracting $1$ gives $2n=60$, and dividing by $2$ yields $n=30$. You have conquered the problem not by grinding through algebra, but by understanding the soul of the polynomial.