Mathematics - Binomial Theorem Question with Solution | TestHub

$R=[R]+f=(8+3\sqrt{7}{)}^{20}$ $={}^{20}{C}_0{8}^{20}+{}^{20}{C}_1{8}^{19}(3\sqrt{7})+\dots \dots$, where $0<f<1$ Let $F=(8-3\sqrt{7}{)}^{20}={}^{20}{C}_0{8}^{20}-{}^{20}{C}_1{8}^{19}(3\sqrt{7})+\dots$.. where $0<F<1$ $\begin{array}{rlr}& \because [R]+f+F=2\left[{}^{20}{C}_0\cdot 8^{20}+{}^{20}{C}_2{8}^{18}(3\sqrt{7}{)}^2+\dots \dots \right]=\text{ an even integer }& \therefore [R]=\text{ an even integer }-1=\text{ an odd integer }\end{array}$ Also, $R-[R]=f=1-F=1-(8-3\sqrt{7}{)}^{20}=1-\frac{1}{(8+3\sqrt{7}{)}^{20}}$ Again $RF=1\Rightarrow R(1-f)=1\Rightarrow R\{1-R+[R]\}=1$ Therefore $R+R[R]=1+R^2$