Animated Solution for Mathematics - Basic Mathematics: Use method of mathematical induction $2.7^n+3.5^n-5$ is divisible by 24 for all $n>0$

Proving Divisibility

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving a problem; we are witnessing the clockwork precision of mathematics.

We are tasked with proving that the expression $P(n)=2\cdot 7^n+3\cdot 5^n-5$ is always divisible by $24$ for any positive integer $n$. This might look like a daunting wall of numbers, but through the lens of Mathematical Induction, it becomes a beautiful, rhythmic dance.

The First Domino

Every great journey begins with a single step. In induction, that is our base case. We test $n=1$ to see if our hypothesis holds ground.

Substituting $n=1$ into our expression, we get:

Calculating this, we find $14+15-5=24$. Since $24$ is clearly divisible by $24$, the first domino has fallen. The foundation is set.

The Inductive Leap

Now, we enter the heart of the logic. We assume the statement is true for some arbitrary integer $k$.

We write this as: where $m$ is an integer. This is our Inductive Hypothesis. We are essentially saying, "If the pattern holds for $k$, what does that force upon $k+1$?"

The Algebraic Alchemy

To prove the statement for $n=k+1$, we look at $P(k+1)=2\cdot 7^{k+1}+3\cdot 5^{k+1}-5$. Using the laws of indices, we rewrite this as:

Here is where the magic happens. We want to force $P(k)$ to appear inside this expression. We know $P(k)=2\cdot 7^k+3\cdot 5^k-5$.

If we multiply this by $7$, we get:

Notice how the first term matches our $P(k+1)$ perfectly! By substituting this back, we perform a bit of algebraic surgery:

Simplifying the remainder, we get:

The Final Cancellation

We are almost there. We know $7\cdot P(k)$ is $7(24m)$, which is divisible by $24$. Now, look at the remainder: $-6(5^k-5)$. We need this to be divisible by $24$ as well.

Recall the identity $x^n-1=(x-1)(x^{n-1}+...+1)$. For $x=5$, $5^k-5=5(5^{k-1}-1)$. Since $5-1=4$, the term $(5^{k-1}-1)$ is a multiple of $4$. Let $5^k-5=4q$.

Substituting this back, we get:

Because $7m-q$ is an integer, $P(k+1)$ is divisible by $24$. We have successfully bridged the gap from $k$ to $k+1$.

By the Principle of Mathematical Induction, the statement holds for all $n>0$. You have conquered the problem, not by brute force, but by understanding the underlying structure of the numbers themselves. Keep this clarity, and no problem will ever be too complex for you.