Animated Solution for Mathematics - Binomial Theorem: Prove that $\frac{n^7}{7}+\frac{n^5}{5}+\frac{2n^3}{3}-\frac{n}{105}$ is an integer for every positive integer $n$.

Unlocking the Divisibility Mystery

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a beautiful algebraic structure. You are looking at the expression $E=\frac{n^7}{7}+\frac{n^5}{5}+\frac{2n^3}{3}-\frac{n}{105}$.

At first glance, it looks like a chaotic mess of fractions. How could such a collection of disparate denominators ever result in a clean, whole integer? It feels counterintuitive, doesn't it?

But in mathematics, chaos is often just order that we haven't yet deciphered.

The Unified Vision

Our first step is to bring order to this chaos. We have denominators of $7,5,3,$ and $105$. The least common multiple is clearly $105$.

Let us rewrite our expression $E$ under this single, unified denominator. By multiplying the terms appropriately, we transform $E$ into a single polynomial $P(n)$:

Now, the problem has shifted. We no longer need to worry about fractions. Our goal is simply to prove that the numerator, $15n^7+21n^5+70n^3-n$, is always divisible by $105$ for any positive integer $n$.

This is the heart of the challenge.

The Foundation

Every great structure needs a foundation. In the language of induction, this is our base case. Let us test $n=1$.

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

Since $105$ is divisible by $105$, our base case holds! We have successfully planted our feet on the first rung of the ladder. Now, we must climb.

The Inductive Leap

We assume that the statement is true for some arbitrary integer $k$. That is, we assume $P(k)$ is an integer. Now, we must prove that $P(k+1)$ is also an integer.

The most elegant way to do this is to examine the difference $D=P(k+1)-P(k)$. If we can show that $D$ is an integer, then since $P(k)$ is an integer, $P(k+1)=P(k)+D$ must also be an integer.

This is the 'Inductive Leap'. Let us write out the difference $D$:

The Binomial Magic

This looks intimidating, but let us breathe. We use the Binomial Theorem to expand these terms. Remember, when we subtract $k^7$ from $(k+1{)}^7$, the leading $k^7$ term vanishes.

We are left with the remaining terms of the expansion:

Now, substitute these back into our expression for $D$. Watch closely as we multiply by the coefficients $15,21,$ and $70$:

Look at the coefficients! $105k^6$, $105k^4$, and $210k^2$ are all clearly multiples of $105$. What about the constants?

We have $15+21+70-1=105$. Every single term in the numerator of $D$ is a multiple of $105$. Therefore, $D$ is an integer.

We have shown that $P(1)$ is an integer, and that if $P(k)$ is an integer, then $P(k+1)$ must also be an integer. By the principle of mathematical induction, the expression is an integer for all positive integers $n$.

You have just navigated a complex proof by breaking it down into manageable, logical steps. This is the essence of JEE Advanced physics and mathematics—not brute force, but elegant, structured reasoning. Keep this mindset, and you will conquer any problem that comes your way.