Question

Let $R={(6\sqrt{6}+14)}^{2n+1}$ and $f=R-\left[R\right]$, where [.] denotes the greatest integer function. Then find the value of $Rf$ is $R{f}^{\prime }$. $n\in N$.

Answer 1

$f=R-\left[R\right]\Rightarrow R=\left[R\right]+f$

$\Rightarrow {(6\sqrt{6}+14)}^{2n+1}=\left[R\right]+f,$ where $\left[R\right]$ is integer and $0\le f<1.$

Now, let $f^{\prime }={(6\sqrt{6}-14)}^{2n+1},0<f^{\prime }<1$

Also $\left[R\right]+f-f^{\prime }={(6\sqrt{6}+14)}^{2n+1}-{(6\sqrt{6}-14)}^{2n+1}$

$=2\left\{{{}^{2n+1}C}_1{\left(6\sqrt{6}\right)}^{2n}\right(14)+{{}^{2n+1}C}_3{\left(6\sqrt{6}\right)}^{2n-2}{\left(14\right)}^3\}$

$=2($ Integer$)=2k$ $(k\in N)\Rightarrow$ even integer

Hence $f-f^{\prime }=$ even integer$-\left[R\right],$ but $-1<f-f^{\prime }<1$

$\because R.H.S$ is integer, hence $L.H.S$ is also integer.

Therefore $f-f^{\prime }=0$$\Rightarrow f=f^{\prime }$

Hence $Rf=R{f}^{\prime }$