Question

If $n$ be a positive integer,and $(7+4\sqrt{3}{)}^n=p+\beta$ where $p$ is a positive integer and $\beta$ is a proper fraction, then value of $(1-\beta )(p+\beta )$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Given
$p+\beta =(7+4\sqrt{3}{)}^n$
$\beta <1$
Assuming
$f=(7-4\sqrt{3}{)}^n$
Clearly,
$f<1$
Now,
$p+\beta +f=(7-4\sqrt{3}{)}^n+(7+4\sqrt{3}{)}^n\Rightarrow p+\beta +f=2[7^n{+}^n{C}_2\cdot 7^{n-2}\cdot (4\sqrt{3}{)}^2+\cdots \cdots \cdots ]\Rightarrow p+\beta +f=2k,k\in N\Rightarrow \beta +f=2k-p\Rightarrow \beta +f=\text{Integer}$
As $\beta +f<2$, so
$\Rightarrow \beta +f=1$

So,
$(p+\beta )(1-\beta )=(p+\beta )\left(f\right)\Rightarrow (p+\beta )(1-\beta )=(7+4\sqrt{3}{)}^n\times (7-4\sqrt{3}{)}^n\Rightarrow (p+\beta )(1-\beta )=(49-16\times 3{)}^n\therefore (p+\beta \left)\right(1-\beta )=1$