Let-a-n-be-a-sequence-such-that-a-1-2-a-n-1-3a-n-4-2a-n-3-n-1-find-a-n-

Question Number 98280 by I want to learn more last updated on 12/Jun/20

Let {a_{n}} be a sequence such that a_{1} = 2,

a_{n + 1} = \frac{3 a_{n} + 4}{2 a_{n} + 3}, n ⩾ 1, find a_{n}

Answered by mr W last updated on 12/Jun/20

a_{n + 1} = \frac{3 a_{n} + 4}{2 a_{n} + 3}

a_{n + 1} + A = \frac{3 a_{n} + 4}{2 a_{n} + 3} + A = \frac{(3 + 2 A) a_{n} + (4 + 3 A)}{2 a_{n} + 3}

a_{n + 1} + B = \frac{3 a_{n} + 4}{2 a_{n} + 3} + B = \frac{(3 + 2 B) a_{n} + (4 + 3 B)}{2 a_{n} + 3}

\frac{a_{n + 1} + A}{a_{n + 1} + B} = \frac{3 + 2 A}{3 + 2 B} \times \frac{a_{n} + \frac{4 + 3 A}{3 + 2 A}}{a_{n} + \frac{4 + 3 B}{3 + 2 B}}

s e t A = \frac{4 + 3 A}{3 + 2 A}

\Rightarrow A^{2} = 2

\Rightarrow A = \sqrt{2}

\Rightarrow B = - \sqrt{2}

\frac{3 + 2 A}{3 + 2 B} = \frac{3 + 2 \sqrt{2}}{3 - 2 \sqrt{2}} = {(3 + 2 \sqrt{2})}^{2}

w i t h b_{n} = \frac{a_{n} + \sqrt{2}}{a_{n} - \sqrt{2}}

b_{n + 1} = {(3 + 2 \sqrt{2})}^{2} b_{n} \leftarrow G . P .

\Rightarrow b_{n} = b_{1} {(3 + 2 \sqrt{2})}^{2 (n - 1)}

b_{1} = \frac{2 + \sqrt{2}}{2 - \sqrt{2}} = 3 + 2 \sqrt{2}

\Rightarrow b_{n} = {(3 + 2 \sqrt{2})}^{2 n - 1}

a_{n} = \frac{\sqrt{2} (b_{n} + 1)}{b_{n} - 1} = \sqrt{2} (1 + \frac{2}{b_{n} - 1})

\Rightarrow a_{n} = \sqrt{2} [1 + \frac{2}{{(3 + 2 \sqrt{2})}^{2 n - 1} - 1}]

w e g e t :

a_{1} = 2

a_{2} = \frac{10}{7}

a_{3} = \frac{58}{41}

a_{4} = \frac{338}{239}

Commented by I want to learn more last updated on 12/Jun/20

Great sir, thanks