Animated Solution for Mathematics - Sequence and Series: Use mathematical Induction to prove : If $n$ is any odd positive integer, then $n(n^2-1)$ is divisible by 24.

A Journey into Number Theory

Welcome, my dear student. Today, we are going to embark on a journey through one of the most beautiful corners of mathematics: Number Theory, specifically using the power of Mathematical Induction.

We are tasked with proving that for any odd positive integer $n$, the expression $n(n^2-1)$ is always divisible by $24$. It might look like a simple algebraic expression, but beneath the surface lies a deep, rhythmic structure of numbers.

The Transformation

When we face a problem involving odd integers, our first instinct should be to bring order to the chaos. We know that any odd positive integer $n$ can be written in the form $n=2m-1$, where $m\in \{1,2,3,\dots \}$.

By making this substitution, we are not just changing variables; we are mapping the infinite set of odd numbers onto the clean, sequential set of natural numbers. Let us define our function as:

The Simplification

Now, let us peel back the layers of this expression. Expanding the squared term, we get $(2m-1{)}^2=4m^2-4m+1$.

Substituting this back into $P(m)$, we see a delightful cancellation: the $+1$ and $-1$ vanish, leaving us with $P(m)=(2m-1)(4m^2-4m)$. If we factor out $4m$ from the second term, we arrive at the elegant form:

This is the engine of our proof.

The Inductive Heart

Mathematical Induction is like a row of falling dominoes. First, we must tip the first one.

For $m=1$, we have $P(1)=4(1)(1-1)(2(1)-1)=0$. Since $0$ is divisible by $24$, our base case is rock solid.

Next, we assume the hypothesis: suppose for some arbitrary integer $k$, $P(k)=4k(k-1)(2k-1)$ is divisible by $24$. We write this as $P(k)=24\lambda$ for some integer $\lambda$. This is our anchor.

The "Aha!" Moment

Now, we look at the target: $P(k+1)$. We substitute $k+1$ into our formula to get:

To prove this is also divisible by $24$, we look at the difference $P(k+1)-P(k)$.

Expanding both, we find $P(k+1)=8k^3+12k^2+4k$ and $P(k)=8k^3-12k^2+4k$. When we subtract them, the $8k^3$ and $4k$ terms vanish into thin air, leaving us with:

This is the magic! We can rewrite this as $P(k+1)=P(k)+24k^2$. Since $P(k)$ is divisible by $24$ (by our hypothesis) and $24k^2$ is clearly a multiple of $24$, their sum must also be divisible by $24$.

We have successfully shown that if the statement holds for $k$, it must hold for $k+1$. By the Principle of Mathematical Induction, the proof is complete.

You have just witnessed how a seemingly complex divisibility problem collapses into a simple, elegant truth. Remember, in JEE Advanced, it is rarely about brute force; it is about finding the right transformation and letting the algebra reveal the underlying beauty. Keep practicing, and keep falling in love with the logic!