Think logically not mathematically [part-31]

Number Theory Level 3

$a_{n} = \frac{1}{\sqrt{5}}\left[ \left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$ Consider a sequence of real numbers $\{a_n\}$ as given above, where $n$ is a non-negative integer . Then which of the following are correct? \[\begin{array} {} \text{A)} & a_{5} & = & 5 \\ \text{B)} & a_{12} & = & 12^2 \\ \text{C)} & a_{10} & = & 55 \\ \text{D)} & a_{2009} + a_{2010} & = & 2 \ a_{2011} \end{array} \]

A, C, D are correct B, C, D are correct A, B, C are correct All are correct A, D are correct C, D are correct A, B are correct B, D are correct

2 solutions

Chew-Seong Cheong
Jun 25, 2016

$\displaystyle a_n = \frac 1{\sqrt 5} \left[ \left(\color{#3D99F6}{\frac {1+\sqrt 5}2} \right)^n - \left(\color{#D61F06}{\frac {1-\sqrt 5}2} \right)^n \right]$ is the Binet's formula $\displaystyle F_n = \frac {\color{#3D99F6}{\varphi}^n - \color{#D61F06}{\psi}^n}{\sqrt 5}$ . Therefore, $a_n$ is the $n$ th Fibonacci number $F_n$ . And $a_5 = F_5 = 5$ , $a_{12} = F_{12} = 144 = 12^2$ and $a_{10} = F_{10} = 55$ but $a_{2009} + a_{2010} =$ $F_{2009} + F_{2010} =$ $F_{2011} =$ $a_{2011} \ne$ $2 \ a_{2011}$ . Therefore, $\boxed{\text{A, B, C are correct}}$ .

Nice solution sir......

sudoku subbu - 4 years, 11 months ago

Prabhav Jain
Jul 3, 2016

Since D is obviously wrong, and C is correct therefore A, B, and C are correct. Think logically, not mathematically.

Why is it 'obviously' wrong?

Arulx Z - 4 years, 9 months ago

Think logically not mathematically [part-31]

2 solutions

0 pending reports