Animated Solution for Mathematics - Binomial Theorem: If $p$ be a natural number then prove that $p^{n+1}+(p+1{)}^{2n-1}$ is divisible by $p^2+p+1$ for every positive integer $n$.

The expression provided is $P(n)=p^{n+1}+(p+1{)}^{2n-1}$. We aim to prove that this expression is always divisible by $p^2+p+1$ for all positive integers $n$.

In the world of competitive mathematics, chaos is often just order waiting to be discovered. Let us embark on this journey of proof using the Principle of Mathematical Induction.

Every great structure needs a solid foundation. In induction, that is our base case. We test the smallest possible value, $n=1$.

When we substitute $n=1$ into our expression, we get:

Simplifying this, we find:

Since $P(1)$ is exactly $p^2+p+1$, it is clearly divisible by $p^2+p+1$. The foundation is set, and the dominoes are ready to fall.

Now, we step into the heart of the proof. We assume the statement is true for some arbitrary integer $k$.

This means we assume $P(k)=p^{k+1}+(p+1{)}^{2k-1}$ is a multiple of $p^2+p+1$. Mathematically, we write: where $m$ is some integer.

This is our 'master key.' We isolate $p^{k+1}$ to obtain: Keep this equation close; it is the secret weapon we will use to conquer the next step.

Now, we look at the case for $n=k+1$. We want to show that $P(k+1)$ is also a multiple of our divisor.

Writing it out, we get:

We decompose this expression as follows:

Now, we substitute our master key into this expression:

We distribute the $p$ and expand the $(p+1{)}^2$ term:

Next, we group the terms containing $(p+1{)}^{2k-1}$:

Simplifying the bracket, $2p-p$ gives us $p$, leaving us with:

We can now factor out the $(p^2+p+1)$ term completely:

Since $P(k+1)$ is a multiple of $p^2+p+1$, the proof is complete by the Principle of Mathematical Induction. The expression is divisible by $p^2+p+1$ for all $n\in {\mathbb{Z}}^{+}$.