Question

If n is a positive integer and $(5\sqrt{5}+11{)}^{2n+1}=I+f$ where I is an integer and 0 < f < 1 then

• A. I is an even integer
• B. $(I+f{)}^2$ is divisible by $2^{2n+1}$
• C. I is divisible by 22
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A I is an even integer
B I is divisible by 22
C $(I+f{)}^2$ is divisible by $2^{2n+1}$

$(5\sqrt{5}+11{)}^{2n+1}=I+F$

Let$(5\sqrt{5}-11{)}^{2n+1}=F$

Now, ${(a+b)}^n={{}^{n}C}_0{a}^n+{{}^{n}C}_1{a}^{n-1}b+{{}^{n}C}_2{a}^{n-2}{b}^2+..........{{}^{n}C}_n{b}^n$

$\Rightarrow {(a-b)}^n={{}^{n}C}_0{a}^n-{{}^{n}C}_1{a}^{n-1}b+{{}^{n}C}_2{a}^{n-2}{b}^2-..........{{}^{n}C}_n{b}^n$

$\Rightarrow {(a+b)}^n-{(a-b)}^n=2[{{}^{n}C}_1{a}^{n-1}b+{{}^{n}C}_3{a}^{n-3}{b}^3+.......]$

$\therefore$ Putting $a=5\sqrt{5};b=11$ and $n=2n+1$ , we get

$\Rightarrow (5\sqrt{5}+11{)}^{2n+1}-(5\sqrt{5}+11{)}^{2n-1}=(I+F)-F$

$=2\left[{{}^{2n+1}C}_1(5\sqrt{5}{)}^{2n}\right(11{)}^1+{{}^{2n+1}C}_3{\left(5\sqrt{5}\right)}^{2n-2}({11)}^3....]$

$\Rightarrow I+F-F=2K,$ $K\in I$

$\Rightarrow I=2K$

$\therefore$ $I$ is an even integer.

Also, on taking 11 common, we get,

$\Rightarrow I=22[{{}^{2n+1}C}_1{\left(5\sqrt{5}\right)}^{2n}+{{}^{2n+1}C}_3{\left(5\sqrt{5}\right)}^{2n-2}{11}^2]$

$\Rightarrow I=22m;m\in$ I

$\therefore I$ is divisible by $22$

$\Rightarrow {(I+F)}^2=[{(5\sqrt{5}+{11)}^{2n+1}]}^2$ = $[{(5\sqrt{5}+{11)}^2]}^{2n+1}$

$={[125+121+110\sqrt{5}]}^{2n+1}$

$={[246+110\sqrt{5}]}^{2n+1}$

$\Rightarrow {(I+F)}^2=2^{2n+1}{[123+55\sqrt{5}]}^{2n+1}$

$={(I+F)}^2$ is divisible by $2^{2n+1}$

Hence, option : A,B,C.