Animated Solution for Mathematics - Binomial Theorem: The number of elements in the set $\{n\in \{1,2,3,\dots ,100\}:(11{)}^n>(10{)}^n+(9{)}^n\}$ is ____.

Welcome, warriors of mathematics! Today, we are going to dismantle a problem that, at first glance, looks like a brute-force nightmare.

Imagine you are staring at the inequality $11^n>10^n+9^n$ for $n$ ranging from $1$ to $100$. If you try to calculate these powers, you will be lost in a sea of digits before you even reach $n=10$.

But here is the secret: JEE Advanced problems are rarely about calculation; they are about pattern recognition. Let us embark on this journey of simplification.

Look at the bases: $11$, $10$, and $9$. Do you see the hidden harmony? They are perfectly symmetric around $10$.

We have $11=10+1$ and $9=10-1$. This is not a coincidence; it is an invitation to use the Binomial Theorem.

Let us rewrite our inequality by moving $9^n$ to the left:

Now, substitute our symmetric bases:

Suddenly, the problem transforms from a daunting exponential inequality into a beautiful algebraic expansion.

Now, let us invoke the Binomial Theorem. We know that:

When we expand $(10+1{)}^n$ and $(10-1{)}^n$, something magical happens. For $(10+1{)}^n$, all terms are positive. For $(10-1{)}^n$, the terms alternate in sign.

When we subtract the second from the first, the even-positioned terms (where $r$ is even) cancel out entirely! The odd-positioned terms (where $r$ is odd) double up.

We are left with:

This is the heart of the problem. We have simplified the left-hand side into a sum of positive terms.

Let us look at the first term of this expansion: $2\cdot {}^n{C}_1\cdot 10^{n-1}$. Since ${}^n{C}_1=n$, this term is simply $2n\cdot 10^{n-1}$.

Our inequality now looks like this:

If we can prove that the first term alone is greater than or equal to $10^n$, then the entire inequality must hold true because the remaining terms are all positive.

So, we set:

Dividing both sides by $10^{n-1}$, we get $2n\ge 10$, which simplifies beautifully to $n\ge 5$.

This is the moment of truth. For any $n\ge 5$, the inequality is guaranteed to be true.

But what about the boundary? We must check $n=4$. Calculating directly: $11^4=14641$, while $10^4+9^4=10000+6561=16561$.

Clearly, $14641<16561$, so $n=4$ fails. Since the gap only widens as $n$ decreases, $n=1,2,3$ will also fail.

We have found our range: $n$ must be in the set $\{5,6,7,\dots ,100\}$. To find the count, we take the total of $100$ numbers and subtract the $4$ invalid cases:

There you have it! Through the power of binomial expansion and logical deduction, we have conquered the problem without ever needing a calculator. Keep this spirit of inquiry alive, and you will solve any problem that comes your way.