Question

If ${(4+\sqrt{15})}^n=I+f$, where $n$ is an odd natural number, $I$ is an integer and $0<f<1$, then

• A. $I$ is an odd integer
• B. $I$ is an even integer
• C. $(I+f)(1-f)=1$
• D. $1-f=(4-\sqrt{15}{)}^n$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $I$ is an odd integer
C $(I+f)(1-f)=1$
D $1-f=(4-\sqrt{15}{)}^n$

$(4+\sqrt{15}{)}^n=I+f-----(1)$

$(4+\sqrt{15}{)}^n+(4-\sqrt{15}{)}^n$

$=2(4^n+{}^n{C}_2{4}^{n-2}(\sqrt{15}{)}^2+{}^n{C}_4{4}^{n-4}(\sqrt{15}{)}^4+...{}^n{C}_n{4}^1(\sqrt{15}{)}^{n-1})$

The above expression is an integer {since there is no fractional or irrational numbers involved}.

Hence, it has to be equal to.

$=I+1$ (Since, $0<(4-\sqrt{15}{)}^n\le 1$)

$(4+\sqrt{15}{)}^n+(4-\sqrt{15}{)}^n=I+1-------\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right),$ we get

$\Rightarrow (4-\sqrt{15}{)}^n=1-f$

$(1-f)(I+f)=(16-15{)}^n$

$=1$ ...(i)

$I+1=2(4^n+{}^n{C}_2{4}^{n-2}(\sqrt{15}{)}^2+{}^n{C}_4{4}^{n-4}(\sqrt{15}{)}^4+...{}^n{C}_n{4}^1(\sqrt{15}{)}^{n-1})$

Hence, $I+1$ is an even number$\Rightarrow$ $I$ is an odd number.