Let u_n = frac{1}{sqrt{5}} left[ left( frac{1 | vedclass

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

Question

(A) The given expression is the Binet's formula for the $n$-th Fibonacci number,where $u_n = F_n$.Let $\alpha = \frac{1 + \sqrt{5}}{2}$ and $\beta = \

Vedclass · Accepted Answer

$u_{n+1} = u_n + u_{n-1}$

Vedclass · Answer

(A) The given expression is the Binet's formula for the $n$-th Fibonacci number,where $u_n = F_n$.Let $\alpha = \frac{1 + \sqrt{5}}{2}$ and $\beta = \frac{1 - \sqrt{5}}{2}$.Then $u_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}$.We know that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2 - x - 1 = 0$,so $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$.For $n \ge 1$,$u_{n+1} = \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}}$.$u_n + u_{n-1} = \frac{1}{\sqrt{5}} [(\alpha^n - \beta^n) + (\alpha^{n-1} - \beta^{n-1})] = \frac{1}{\sqrt{5}} [\alpha^{n-1}(\alpha + 1) - \beta^{n-1}(\beta + 1)]$.Since $\alpha + 1 = \alpha^2$ and $\beta + 1 = \beta^2$,we get $u_n + u_{n-1} = \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}} = u_{n+1}$.Thus,the recurrence relation is $u_{n+1} = u_n + u_{n-1}$.

Let $u_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$,for $n = 0, 1, 2, ...$. Then which of the following is true?

Similar Questions

If $\alpha, \beta$ are natural numbers such that $100^{\alpha} - 199\beta = (100)(100) + (99)(101) + (98)(102) + \ldots + (1)(199)$,then the slope of the line passing through $(\alpha, \beta)$ and the origin is:

The value of the expression $1.(2 - \omega )(2 - {\omega ^2}) + 2.(3 - \omega )(3 - {\omega ^2}) + ....... + (n - 1).(n - \omega )(n - {\omega ^2}),$ where $\omega$ is an imaginary cube root of unity,is

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

Find the sum to $n$ terms of the series $3 \times 8 + 6 \times 11 + 9 \times 14 + \dots$

If $S_{1}, S_{2}, S_{3}$ are the sum of first $n$ natural numbers,their squares,and their cubes,respectively,show that $9 S_{2}^{2} = S_{3}(1 + 8 S_{1})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module