Question

If I is the integral part and f the fractional part of $(5\sqrt{2}+7{)}^{2n+1}$ then prove that $f(I+f)$ = 1

Answer 1

It is exactly as part (a).
$(5\sqrt{2}+7{)}^{2n+1}$ = I + f o < f < 1
$(5\sqrt{2}-7{)}^{2n+1}$ = f' o < f' < 1
Subtracting I + f - f' = Even integer
It will be possible only when f - f' = 0 as both lie between 0 and 1.
$\therefore$ f = f'
$\therefore$ f(I + f) = f' (I + f)
$=\left[\right(5\sqrt{2}-7\left)\right(5\sqrt{2}+7){]}^{2n+1}$
$=[50-49{]}^{2n+1}=1.$