Question

please solve this answer i have alot of qiestions

Answer 1

$let \: x = 4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + .... \infty } } } \\ then \\ x = 4 + \frac{1}{x} \\ {x}^{2} - 4x - 1 = 0 \\ x = \frac{4 + \: \: or \: \: - \sqrt{ {( - 4)}^{2} - 4 \times 1 \times - 1} }{2 \times 1} \\ x = 2 + \sqrt{5} \: \: or \: \: 2 - \sqrt{5} \\ since \: \: - ve \: \: value \: \: is \: \: not \: \: admissible \\ 4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + .... \infty } } } = 2 + \sqrt{5}$