Question

Match the elements of List 1 with elements of List 2:
$n\in I^{+}$

Answer 1

A) Let ${(\sqrt{2}+1)}^{2n}=I+F$ where $I$ is some integer and $0\le F<1$.
Then we have
$\sqrt{2}-1=\frac{1}{\sqrt{2}+1}$ $\Rightarrow 0<\sqrt{2}-1<1$
$\Rightarrow 0<f={(\sqrt{2}-1)}^{2n}<1$
Also ,
$I+F+f={(\sqrt{2}-1)}^{2n}+{(\sqrt{2}+1)}^{2n}$
$=2[{{}^{2n}C}_0{2}^n+{{}^{2n}C}_2{2}^{n-1}+.....+{{}^{2n}C}_{2n}]$
$=$ an even integer, say $4m+2$
Therefore
$\Rightarrow F+f=4m+2-I$
As $0\le f<1$ and $0<F<1$
thus,
$0<f+F<2$ $\Rightarrow f+F=1$
$\Rightarrow I=4m+1$
which is of the form $4k+1$
B)
Next, as $0<2-\sqrt{3}=\frac{1}{2+\sqrt{3}}<1$
Thus
${(2-\sqrt{3})}^{2n}=0$
C) Let $F_1={(2-\sqrt{3})}^{2n}$
Also let ${(2-\sqrt{3})}^{2n}=I_1+f_1$, where $0\le f_1<1$
We have
$I_1+f_1+F_1={(2-+\sqrt{3})}^{2n}+{(2-\sqrt{3})}^{2n}$
As above $f_1+F_1=1$, therefore $I_1$ is of the form $4k+1$
D) $0<\sqrt{2}-1<1,{(\sqrt{2}-1)}^{2n}=0$