Question

The sequence {S_n} defined by \(S_{n+1}=\sqrt{7+S_n}, \ S_1 = \sqrt{7}\) converges to
1. Positive root of x² - x - 7 = 0
2. Negative root of x² - x - 7 = 0
3. Positive root of x² + x - 7 = 0
4. Negative root of x² + x - 7 = 0

Answer 1

Correct Answer - Option 1 : Positive root of x² - x - 7 = 0

Given:

\({S_{n + 1}} = \sqrt {7 + {S_n}}\) 

\({S_1} = \sqrt 7 \) 

Analysis:

\({S_2} = \sqrt {7 + \sqrt 7 }\) 

\({S_3} = \sqrt {7 + \sqrt {7 + \sqrt 7 } }\) 

\(\mathop {\lim }\limits_{n \to \infty } {S_{n + 1}} = \sqrt {7 + \sqrt {7 + \sqrt {7 + \ldots } } }\)       ---(1)

Let \(\sqrt {7 + \sqrt {7 + \sqrt {7 + \ldots } } } = x\) 

∴ \(x = \sqrt {7 + x} \)

x² = 7 + x

x² – x – 7 = 0

\(x = \frac{{1 \pm \sqrt {1 + 28} }}{2}\) 

As the series converges to a positive number

∴ Positive roots of x² – x – 7 is the solution.