Animated Solution for Mathematics - Binomial Theorem: Let $R=(5\sqrt{5}+11{)}^{2n+1}$ and $f=R-[R]$, where $[]$ denotes the greatest integer function. Prove that $Rf=4^{2n+1}$.

Imagine you are standing at the base of a massive, daunting mountain. The problem before us is $R=(5\sqrt{5}+11{)}^{2n+1}$, and we are asked to determine its fractional part, $f=R-[R]$.

At first glance, this expression looks like a nightmare. It is an irrational number raised to an odd power. The secret, as with many great challenges in JEE mathematics, is to find the hidden path around it: the conjugate.

Whenever you see an expression like $(a+\sqrt{b}{)}^n$, your mathematical intuition should immediately scream, "Conjugate!" Let us define:

Note that $5\sqrt{5}=\sqrt{125}$, which is slightly larger than $\sqrt{121}=11$. This means $5\sqrt{5}-11$ is a tiny, positive fraction, strictly between $0$ and $1$.

When you raise a number between $0$ and $1$ to any positive power, it remains between $0$ and $1$. Thus, we establish the bedrock of our proof: $0<R^{\prime }<1$.

Now, let us expand both $R$ and $R^{\prime }$ using the Binomial Theorem. For $R=(5\sqrt{5}+11{)}^{2n+1}$, every term in the expansion is positive.

For $R^{\prime }=(5\sqrt{5}-11{)}^{2n+1}$, the terms alternate in sign because of the negative $11$. When we look at the sum $R+R^{\prime }$, the terms involving odd powers of $5\sqrt{5}$ (the irrational parts) cancel out perfectly.

What remains is twice the sum of the terms with even powers of $5\sqrt{5}$. Since $(5\sqrt{5}{)}^2=125$, these terms are all integers. Thus, $R+R^{\prime }$ is an even integer, which we can call $I$.

We know $R=[R]+f$. Substituting this into our equation $R+R^{\prime }=I$, we get $[R]+f+R^{\prime }=I$.

Rearranging gives $f+R^{\prime }=I-[R]$. Since $I$ and $[R]$ are both integers, their difference must be an integer.

We have established that $f+R^{\prime }$ is an integer. Given the bounds $0\le f<1$ and $0<R^{\prime }<1$, we find:

The only integer in the interval $(0,2)$ is $1$. Therefore, $f+R^{\prime }=1$, which means the fractional part is $f=1-R^{\prime }$.

Finally, we consider the product $R\cdot f$. Since $f=1-R^{\prime }$, we have:

Substituting the values:

This simplifies to:

We have reached the summit. The complexity vanishes, leaving behind the result $R\cdot f=R-4^{2n+1}$, where $f=1-(5\sqrt{5}-11{)}^{2n+1}$.