Question

If $R=(6\sqrt{6}+14{)}^{2n+1}$ and f=R-[R], where [.] denotes the greatest integer function, then Rf equals:

• A. 20ⁿ
• B. 20²ⁿ
• C. 20²ⁿ⁺¹
• D. None

Answer 1

The correct option is C

20²ⁿ⁺¹

R = [R] + f = $(6\sqrt{6}+14{)}^{2n+1},0\le f<1$

Let $S=(6\sqrt{6}+14{)}^{2n+1}$ , 0 < S < 1

[R] + f - S = 2 { ${}^{2n+1}{C}_1(6\sqrt{6}{)}^{2n}.14+$

${}^{2n+1}{C}_3(6\sqrt{6}{)}^{2n-3}.{14}^3+....$ } = int eger.

[R] is int eger, so f - S is int eger.

0 $\le f<1,-1<-S<0$

$\Rightarrow$ -1 < f < -S < 1 $\Rightarrow$ f - S = 0

$\therefore$ Rf = RS = $(6\sqrt{6}+14{)}^{2n+1}(6\sqrt{6}-14{)}^{2n+1}$

= ${20}^{2n+1}$