Animated Solution for Mathematics - Binomial Theorem: Prove that $7^{2n}+2^{3n-3}(3^{n-1})$ is divisible by 25 for any natural number $n$.

Unveiling Hidden Divisibility

Welcome, fellow traveler on the path of JEE Advanced mastery. Today, we aren't just solving a problem; we are uncovering a hidden symmetry in numbers.

We are tasked with proving that the expression $P(n)=7^{2n}+2^{3n-3}\times 3^{n-1}$ is always divisible by $25$ for any natural number $n$. It looks intimidating, but remember, complexity is often just simplicity in disguise.

Simplifying the Landscape

Before we charge into the battlefield of induction, let us tidy up our workspace. We have $P(n)=7^{2n}+2^{3n-3}\times 3^{n-1}$.

The first term, $7^{2n}$, is easily seen as $(7^2{)}^n=49^n$.

Now, look at the second term. We can write $2^{3n-3}$ as $2^{3(n-1)}$, which is $8^{n-1}$. Combining this with $3^{n-1}$ using the laws of exponents, we get $(8\times 3{)}^{n-1}=24^{n-1}$.

Our expression has transformed into something much more elegant:

The Foundation (Base Case)

Every great structure needs a solid foundation. We test $n=1$.

Substituting this into our simplified expression, we get $P(1)=49^1+24^{1-1}=49+24^0$.

Since any non-zero number to the power of zero is $1$, we have $49+1=50$. Since $50=2\times 25$, the base case is undeniably true.

The Inductive Leap

Now, we assume the statement holds for some arbitrary integer $k$. We assume $P(k)=49^k+24^{k-1}=25m$, where $m$ is some integer.

This is our 'Inductive Hypothesis.' We rearrange this to isolate $49^k$:

Now, we must prove the statement for $n=k+1$. We look at $P(k+1)=49^{k+1}+24^{(k+1)-1}=49^{k+1}+24^k$.

Breaking the first term down, we have $49^{k+1}=49\times 49^k$. Substituting our hypothesis, we get:

The Grand Finale

This is where the magic happens. We expand the expression: $P(k+1)=49\times 25m-49\times 24^{k-1}+24^k$.

To combine the terms involving $24$, we rewrite $24^k$ as $24\times 24^{k-1}$. Factoring out $24^{k-1}$, we get $(-49+24)\times 24^{k-1}=-25\times 24^{k-1}$.

Our expression now stands as:

Finally, we factor out the $25$:

Since $m$ and $k$ are integers, the term inside the parenthesis is also an integer. We have proven that $P(k+1)$ is a multiple of $25$.

By the Principle of Mathematical Induction, we have conquered the problem. The math didn't just work; it flowed.