Question

Let ${(5+2\sqrt{6})}^n=p+f$ where $nϵN$ and $pϵN$ and $0<f<1$, then the value of $f^2-f+pf-p$ is

• A. a natural number
• B. a negative integer
• C. a prime number
• D. an irrational number

Answer 1

The correct option is B a negative integer

${(5+2\sqrt{6})}^n$can be written in the form${(5+\sqrt{24})}^n$

Now,let $I+f={(5+\sqrt{24})}^n$ ...(1)

As f is the fractional part$.0\le f<1$ ...(2)and

let$f^I={(5-\sqrt{24})}^n.$ .....(3)

$0\le f^I<1$. ...(4)

On adding Eqs.(i)and (iii),we get $I+f+f^I={\sqrt{(5+\sqrt{24})}}^n{\sqrt{5-\sqrt{24}}}^n$

$I+1=2p$=even integer.

$I=2p-1=$odd integer.

From the question$f^2-f+If-I=f^2+If-f-I=f(f+I)-1(f+1)=(f+I)(f-1)=(f+2p-1)(f-1)$ where f is the fractional part and $(f-1)<0$

Hence$f<1.$

As the integral part is greater than the fractional part,fractional part-1=negative\quad integer.

And$(f+2p-1)>0$

Hence,its value is a negative integer.