1-simplify-A-n-1-2-i-3-n-1-2-i-3-n-2-smplify-B-n-1-2-3-n-1-2-3-n-n-integr-natural-

Question Number 49952 by maxmathsup by imad last updated on 12/Dec/18

1) s i m p l i f y A_{n} = \frac{1}{{(2 + i \sqrt{3})}^{n}} + \frac{1}{{(2 - i \sqrt{3})}^{n}}

2) s m p l i f y B_{n} = \frac{1}{{(2 + \sqrt{3})}^{n}} + \frac{1}{{(2 - \sqrt{3})}^{n}}

n i n t e g r n a t u r a l .

Commented by Abdo msup. last updated on 14/Dec/18

1) w e h a v e A_{n} = \frac{{(2 + i \sqrt{3})}^{n} + {(2 - i \sqrt{3})}^{n}}{7^{n}}

= \frac{2 R e ({(2 + i \sqrt{3})}^{n})}{7^{n}} b u t 2 + i \sqrt{3} = \sqrt{7} (\frac{2}{\sqrt{7}} + \frac{i \sqrt{3}}{\sqrt{7}}) = r e^{i θ} \Rightarrow

r = \sqrt{7} a n d c o s θ = \frac{2}{\sqrt{7}}, s i n θ = \frac{\sqrt{3}}{\sqrt{7}} \Rightarrow t a n θ = \frac{\sqrt{3}}{2} \Rightarrow θ = a r c t a n (\frac{\sqrt{3}}{2})

\Rightarrow {(2 + i \sqrt{3})}^{n} = {(\sqrt{7})}^{n} e^{i n a r c t a n (\frac{\sqrt{3}}{2})} \Rightarrow \frac{2 {(\sqrt{7})}^{n} c o s (n a r c t a n (\frac{\sqrt{3}}{2}))}{7^{n}}

= \frac{2}{{(\sqrt{7})}^{n}} c o s (n a r c t a n (\frac{\sqrt{3}}{2})) .

Commented by Abdo msup. last updated on 14/Dec/18

2) w e h s v e B_{n} = \frac{{(2 + \sqrt{3})}^{n} + {(2 - \sqrt{3})}^{n}}{{(2^{2} - {(\sqrt{3})}^{2})}^{n}}

= \sum_{k = 0}^{n} C_{n}^{k} 2^{n - k} {(\sqrt{3})}^{k} + \sum_{k = 0}^{n} C_{n}^{k} 2^{n - k} {(- \sqrt{3})}^{k}

= \sum_{k = 0}^{n} C_{n}^{k} {{(\sqrt{3})}^{k} + {(- \sqrt{3})}^{k}) 2^{n - k}

= \sum_{p = 0}^{[\frac{n}{2}]} C_{n}^{2 p} {3^{p} + 3^{p}) 2^{n - 2 p}

= 2^{n + 1} \sum_{p = 0}^{[\frac{n}{2}]} {(\frac{3}{4})}^{p} C_{n}^{2 p}