The Fibonacci sequence is defined by a_1 = 1, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $a_1 = 1, a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$.Calculating the first few terms:$a_3 = a_2 + a_1 = 1 + 1 = 2$$a_4 = a_3 + a_2 =

Vedclass · Accepted Answer

$1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}$

Vedclass · Answer

(A) Given $a_1 = 1, a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$.Calculating the first few terms:$a_3 = a_2 + a_1 = 1 + 1 = 2$$a_4 = a_3 + a_2 = 2 + 1 = 3$$a_5 = a_4 + a_3 = 3 + 2 = 5$$a_6 = a_5 + a_4 = 5 + 3 = 8$Now,calculating the ratios $\frac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$:For $n = 1: \frac{a_2}{a_1} = \frac{1}{1} = 1$For $n = 2: \frac{a_3}{a_2} = \frac{2}{1} = 2$For $n = 3: \frac{a_4}{a_3} = \frac{3}{2}$For $n = 4: \frac{a_5}{a_4} = \frac{5}{3}$For $n = 5: \frac{a_6}{a_5} = \frac{8}{5}$Thus,the sequence of ratios is $1, 2, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}$.

The Fibonacci sequence is defined by $a_1 = 1, a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$. Find $\frac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$.

Similar Questions

If $x_n = \frac{2n^2 + n + 1}{2n^2 - 3n + 2}$,then $\sum_{r=1}^n \left[ \left( \prod_{i=1}^r x_i \right) - 2\sum_{i=1}^r (2i - 1) \right]$ is equal to

What is the sum of the infinite terms of the geometric progression $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{2 - \sqrt{2}}, \frac{1}{2}, \dots$?

If $a, x$ are real numbers and $|a| < 1, |x| < 1$,then $1 + (1+a)x + (1+a+a^2)x^2 + \dots \infty$ is equal to

The $20^{th}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$ will be

The sum to $n$ terms of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module