Question

Let \(n\) be a positive integer. Show that the smallest integer greater than \((\sqrt{3} + 1)^{2n}\)is divisible by \(2^{n+1}.\)

Hint:
Prove that \(\lceil (\sqrt{3}+1)^{2n} \rceil = (\sqrt{3}+1)^{2n} + (\sqrt{3}-1)^{2n}.\)

Answer 1

you can use mathematical induction with binomial expansion.

Answer 2

Id probably say make a recurrence relation

Answer 3

A number can't be equal to itself plus a nonzero number...

Answer 4

Isn't that\[\lceil (\sqrt{3}+1)^{2n} \rceil = (\sqrt{3}+1)^{n} + (\sqrt{3}-1)^{n}.\]

Answer 5

\[\exists ! 0.abcd.... ~s.t.\]
\[\lceil{(\sqrt{3}+1)^{2n}}\rceil=(\sqrt{3}+1)^{2n}+0.abcd.....\]
\[But~ 0<(\sqrt{3}-1)^{2n}< 1\forall n\in \mathbb{Z}^+\]
By binomial expansion:
\[\small (\sqrt{3}+1)^{2n}=^{2n}C_0 (\sqrt{3})^{2n}+^{2n}C_1 (\sqrt{3})^{2n-1}+\ldots+^{2n}C_{2n-1} (\sqrt{3})^{1}+^{2n}C_{2n} 1 ~~(2n+1~Terms)\]
\[\small (\sqrt{3}-1)^{2n}=^{2n}C_0 (\sqrt{3})^{2n}-^{2n}C_1 (\sqrt{3})^{2n-1}+\ldots-^{2n}C_{2n-1} (\sqrt{3})^{1}+^{2n}C_{2n} 1 ~~(2n+1~Terms)\]
\[\small \therefore (\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}=2 \left[^{2n}C_0 (\sqrt{3})^{2n}+^{2n}C_{2} (\sqrt{3})^{2n-2}+\ldots+^{2n}C_{2n-2} (\sqrt{3})^{2}+^{2n}C_{2n} 1\right] \]
Since all the terms have even exponents of \(\sqrt{3}\) this will be an integer...
\[\therefore \lceil{(\sqrt{3}+1)^{2n}}\rceil=(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}\]
By mathematical induction we can show that..
\[2^{n+1}|\lceil{(\sqrt{3}+1)^{2n}}\rceil\]

Answer 6

suppose \(a=a+b\) where \(b\ne 0\). Then by cancelation \(b=0\) a contradiction.

:)

Answer 7

@zzr0ck3r you've forgotten the ceiling... :)
for example \(\lceil 2.5\rceil =3\)

Answer 8

Ahh, sorry on my phone and I thought it was just \([\). lol. I was going nuts...