Question

Suppose that $n$ is a natural number and $I,F$ are respectively the integral part and fractional part of $(7+4\sqrt{3}{)}^n$. Then show that
i) $I$ is an odd integer
ii) $(I+F)(1-F)=1$.

Answer 1

Let $(7+4\sqrt{3}{)}^n=I+F$

$\therefore I+F=7^n+7^{n-1}\left(4\sqrt{3}{)}^1{(}^n{C}_1\right)+7^{n-2}(4\sqrt{3}{)}^2{}^n{C}_2.....+(4\sqrt{3}{)}^n$

We know that $4\sqrt{3}$ is a fractional term and only even powers of $4\sqrt{3}$ will be non fractional because $\sqrt{3}$ will be squared in those.

$\therefore$ All the terms with even powers of $4\sqrt{3}$ are integers and all terms with odd powers of $4\sqrt{3}$ are fractional

$\therefore I+F=7^n+7^{n-2}\left(4\sqrt{3}{)}^2{(}^n{C}_2\right)+7^{n-4}\left(4\sqrt{3}{)}^4{(}^n{C}_4\right)+....{.7}^{n-1}(4\sqrt{3}{)}^n{C}_1+7^{n-3}(4\sqrt{3}{)}^3{}^n{C}_3+....$

Separating the integer and fractional parts

$I=7^n+7^{n-2}(4\sqrt{3}{)}^2$ ${}^n{C}_2+7^{n-4}(4\sqrt{3}{)}^4{}^{n{C}_4}+....$ ......(i)

$F=7^{n-1}(4\sqrt{3}{)}^1{}^n{C}_1+7^{n-3}(4\sqrt{3}{)}^3{}^n{C}_3+....$ .........(ii)

Now consider I,

$7$ raised to any power will be an odd number -A

All even powers of $4\sqrt{3}$ will be even -B

even term $\times$ odd term$=$even term-C

using A, B and C we can see that all terms in I are even except $7^n$

$I=7^n+7^{n-2}(4\sqrt{3}{)}^2{}^n{C}_2+.....$

any odd term$+$ even term$=$odd term

$\therefore$ I is always odd.