Question

If $p=(8+3\sqrt{7}{)}^n$ and $f=p-\left[p\right],$ where $[\cdot ]$ denotes the greatest integer function, then the value of $p(1-f)$ is

• A. $1$
• B. $2$
• C. $2^n$
• D. $2^{2n}$

Answer 1

The correct option is A $1$

Given $p={(8+3\sqrt{7})}^n$

and $p=\left[p\right]+f$ (where$[.]=$GIF)

Sol. : Let$N={(8-3\sqrt{7})}^n$

$\Rightarrow$ Now, ${(a+b)}^n+{(a-b)}^n=2[{{}^{n}C}_0{a}^n+{{}^{n}C}_2{a}^{n-2}-b^2+...b^n]$

$\Rightarrow$ Take $a=8$ & $b=3\sqrt{7}$

$\therefore p+N=2[{{}^{n}C}_0{8}^n+{{}^{n}C}_2{8}^{n-2}-{\left(3\sqrt{7}\right)}^2+....]$

$=2.K$ (where $k$ &$I$)

Now, $0<f<1$ & $0<N<1$ $\because N={(\sqrt{64}-\sqrt{63})}^n$

$\therefore p+N=\left[p\right]+f+N=2K$

$\left[P\right]$ is always integer & $p+N$ is an integer

$=f+N$ must be an integer.

But, $0<(f+N)<2$ $\to f+N=1$

$\Rightarrow p(1-f)=p.N={(8+3\sqrt{7})}^n{(8-3\sqrt{7})}^n$

$\Rightarrow {(8^2+(3\sqrt{7}{)}^2)}^n=1^n=1.$

Hence, the answer is $1.$