If-a-b-1-a-2-b-2-2-then-a-11-b-11-

Question Number 210591 by Tawa11 last updated on 13/Aug/24

If

a + b = 1

a^{2} + b^{2} = 2

then, a^{11} + b^{11} = ? ?

Answered by mr W last updated on 13/Aug/24

p_{1} = e_{1} = 1

p_{1}^{2} = 2 + 2 e_{2} = 1 \Rightarrow e_{2} = - \frac{1}{2}

p_{n} = e_{1} p_{n - 1} - e_{2} p_{n - 2}

p_{n} - p_{n - 1} - \frac{p_{n - 2}}{2} = 0

r^{2} - r - \frac{1}{2} = 0

\Rightarrow r = \frac{1 \pm \sqrt{3}}{2}

\Rightarrow p_{n} = a^{n} + b^{n} = {(\frac{1 + \sqrt{3}}{2})}^{n} + {(\frac{1 - \sqrt{3}}{2})}^{n}

a^{11} + b^{11} = {(\frac{1 + \sqrt{3}}{2})}^{11} + {(\frac{1 - \sqrt{3}}{2})}^{11} = \frac{989}{32}

Commented by Tawa11 last updated on 13/Aug/24

Wow . Thanks sir .

I appreciate .

Answered by mr W last updated on 13/Aug/24

a^{2} + b^{2} + 2 a b = {(a + b)}^{2} \Rightarrow a b = \frac{1^{2} - 2}{2} = - \frac{1}{2}

{(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)

\Rightarrow a^{3} + b^{3} = 1^{3} - 3 (- \frac{1}{2}) \times 1 = \frac{5}{2}

(a^{3} + b^{3}) (a^{2} + b^{2}) = a^{5} + b^{5} + {(a b)}^{2} (a + b)

\Rightarrow a^{5} + b^{5} = \frac{5}{2} \times 2 - {(- \frac{1}{2})}^{2} \times 1 = \frac{19}{4}

{(a^{3} + b^{3})}^{2} = a^{6} + b^{6} + 2 {(a b)}^{3}

\Rightarrow a^{6} + b^{6} = {(\frac{5}{2})}^{2} - 2 {(- \frac{1}{2})}^{3} = \frac{13}{2}

(a^{5} + b^{5}) (a^{6} + b^{6}) = a^{11} + b^{11} + {(a b)}^{5} (a + b)

\Rightarrow a^{11} + b^{11} = \frac{19}{4} \times \frac{13}{2} - {(- \frac{1}{2})}^{5} \times 1 = \frac{989}{32} ✓

Commented by Tawa11 last updated on 13/Aug/24

Thanks sir .

I appreciate .

Answered by ajfour last updated on 14/Aug/24

a (1 + t) = 1

a^{2} (1 + t^{2}) = 2

a^{11} (1 + t^{11}) = x

t = \frac{b}{a} = 2 a b + 2 = 1 \Rightarrow a b = - \frac{1}{2}

a^{2} t = - \frac{1}{2}

a^{2} {(1 + t)}^{2} = 1

\frac{{(1 + t)}^{2}}{t} = - 2

\Rightarrow t^{2} + 4 t + 4 = 3

t = - 2 \pm \sqrt{3}

a = \frac{1}{1 + t}

a^{11} + b^{11} = \frac{1 + {(- 2 \pm \sqrt{3})}^{11}}{{(- 1 \pm \sqrt{3})}^{11}}

Commented by Tawa11 last updated on 14/Aug/24

Thanks sir .

I appreciate .

Facebook Tweet Pin