Animated Solution for Mathematics - Binomial Theorem: Using mathematical induction prove that for every integer $n\ge 1,(3^{2n}-1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$.

Unmasking the Hidden Truth

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a mathematical investigation. In the world of JEE Advanced, the most dangerous trap is not the complexity of the calculus or the depth of the geometry—it is the assumption that the problem statement is always perfect.

Let us begin by being detectives. We are asked to prove that $3^{2^n}-1$ is divisible by $2^{n+2}$ but not by $2^{n+3}$. If you blindly start induction on the expression $3^{2n}-1$, you will hit a wall.

Let us test $n=3$:

Now, $2^{3+2}=32$. Is $728$ divisible by $32$?

It fails! This is our first victory. We have identified a typo. The expression must be $3^{2^n}-1$. Always verify your terrain before you march.

The Base Case

Every great structure needs a solid foundation. For $n=1$, our expression becomes:

The divisor is $2^{1+2}=2^3=8$. Clearly, $8$ is divisible by $8$.

Now, check the second condition: is it divisible by $2^{1+3}=16$? No, $8$ is not divisible by $16$. The base case holds. We have our starting point.

The Secret of Parity

Now, we assume the statement is true for $n=k$. We write:

Here is where the magic happens. We must define $m$ as an odd integer. Why? Because if $m$ were even, the expression would contain an extra factor of $2$, making it divisible by $2^{k+3}$, which would violate the second part of our proof.

By forcing $m$ to be odd, we lock in the exact power of $2$ that divides our expression. Rearranging this, we get:

Keep this equation close; it is our most powerful weapon.

Algebra in Action

We want to prove the statement for $n=k+1$. Our target is $3^{2^{k+1}}-1$. Using the laws of exponents, we know $2^{k+1}=2^k\cdot 2$.

Thus, $3^{2^{k+1}}=(3^{2^k}{)}^2$. Our expression becomes:

Now, substitute our hypothesis:

Expand this using $(a+b{)}^2=a^2+2ab+b^2$. We get:

The $+1$ and $-1$ vanish, leaving us with:

The Parity Check

We need to show this is equal to $M\cdot 2^{k+3}$ where $M$ is odd. Factor out $2^{k+3}$:

Let $M=m^2\cdot 2^{k+1}+m$. Since $k\ge 1$, $2^{k+1}$ is even, so $m^2\cdot 2^{k+1}$ is even.

Since $m$ is odd, $M=\text{Even}+\text{Odd}=\text{Odd}$. We have done it!

We have shown that $3^{2^{k+1}}-1$ is $M\cdot 2^{k+3}$ with $M$ odd. This means it is divisible by $2^{k+3}$ but not by $2^{k+4}$. The proof is complete. You have navigated the typo, mastered the hypothesis, and conquered the algebra. Keep this analytical mindset, and no JEE problem will ever stand in your way.