Question

Which of the following is true about $(\sqrt{3}+1{)}^{2n}$, where n is a positive integer

• A. The integer just above it is an even number
• B. The integer above it is divisible by
• C. The integer just above it is divisible by 3
• D. The integer just above is divisible by

Answer 1

Answer: (A), (B)

The correct options are
A

The integer just above it is an even number

B

The integer above it is divisible by

$(\sqrt{3}+1{)}^2$ = (4 + $2\sqrt{3}$) = $2(2+\sqrt{3})$

⇒ $(\sqrt{3}+1{)}^{2n}$ = $2^n(2+\sqrt{3}{)}^n$

Consider $(\sqrt{3}-1)$

0 < $(\sqrt{3}-1)$ < 1

⇒ 0 < $(\sqrt{3}-1{)}^n$ < 1

Also, $(\sqrt{3}-1{)}^{2n}$ = $(4-2\sqrt{3}{)}^n$ = $2^n(2-\sqrt{3}{)}^n$

Let $2^n(2+\sqrt{3}{)}^n$ = I + f, where I is the integral part and f is the fractional part.

Let $2^n(2-\sqrt{3}{)}^n$ = f' , 0 < f' < 1

I + f + f' = $(\sqrt{3}+1{)}^{2n}$ + $(\sqrt{3}-1{)}^{2n}$

= $2^n\left[\right(2+\sqrt{3}{)}^n+(2-\sqrt{3}{)}^n]$

= $2^n$[2(${}^n{C}_0$ $2^n$) + ${}^n{C}_2$$2^{n-2}$$(\sqrt{3}{)}^2$ + ............)]

= $2^{n+1}$K, where_k_is_a_positive_integer)

0 < f + f' < 2

For I + f + f' to be an integer f + f' should also be an integer. The only integer value it

can take is 1.

I + f + f' is an even integer ⇒ I is an odd integer

The integer just above $(\sqrt{3}+1{)}^{2n}$ is $2^{n+1}$ K

⇒ A & B