# What is the rule for the sequence 3, 5, 8, 13, 21,... ?

Dec 6, 2016

nth term plus the nth + 1 term:

#### Explanation:

This sequence is the:

nth term plus the nth + 1 term:

$3 + 5 = 8 , 5 + 8 = 13 , 8 + 13 = 21 , 13 + 21 = 34$

This is also called the Fibonacci Series.

Dec 6, 2016

The general term is given by the formula:

${a}_{n} = \left(\frac{3}{2} + \frac{7}{10} \sqrt{5}\right) {\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{n - 1} + \left(\frac{3}{2} - \frac{7}{10} \sqrt{5}\right) {\left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right)}^{n - 1}$

#### Explanation:

The Fibonacci sequence is defined by:

${F}_{0} = 0$

${F}_{1} = 1$

${F}_{n + 2} = {F}_{n} + {F}_{n + 1}$

The first few terms are:

$0 , 1 , 1 , 2 , \textcolor{b l u e}{3 , 5 , 8 , 13 , 21} , 34 , 55 , 89 , 144 , \ldots$

Note that the given sequence starts at ${F}_{4} = 3$, but otherwise follows the same rules.

${a}_{1} = 3$

${a}_{2} = 5$

${a}_{n + 2} = {a}_{n} + {a}_{n + 1}$

In order to find a general formula consider the geometric sequence:

$1 , x , {x}^{2} , \ldots$

If this sequence satisfies the same recursive rule as the Fibonacci sequence then:

${x}^{2} = 1 + x$

So:

$0 = {x}^{2} - x - 1 = {\left(x - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{5}}{2}\right)}^{2} = \left(x - \frac{1}{2} - \frac{\sqrt{5}}{2}\right) \left(x - \frac{1}{2} + \frac{\sqrt{5}}{2}\right)$

Hence: $\text{ "x = 1/2+sqrt(5)/2" }$ or $\text{ } x = \frac{1}{2} - \frac{\sqrt{5}}{2}$

Consider the sequence:

${b}_{n} = A {\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{n - 1} + B {\left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right)}^{n - 1}$

where $A$ and $B$ are any constants.

Notice that any such sequence ${b}_{n}$ will satisfy the recursive rule:

${b}_{n + 2} = {b}_{n} + {b}_{n + 1}$

So if we can find values for $A$ and $B$ such that ${b}_{1} = {a}_{1}$ and ${b}_{2} = {a}_{2}$, then we have a general formula for our sequence.

So we just require:

$3 = {b}_{1} = A {\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{1 - 1} + B {\left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right)}^{1 - 1}$

$\textcolor{w h i t e}{3} = A + B$

$5 = {b}_{2} = A {\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{2 - 1} + B {\left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right)}^{2 - 1}$

$\textcolor{w h i t e}{5} = \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) A + \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) B$

$\textcolor{w h i t e}{5} = \frac{1}{2} \left(A + B\right) + \frac{\sqrt{5}}{2} \left(A - B\right)$

$\textcolor{w h i t e}{5} = \frac{3}{2} + \frac{\sqrt{5}}{2} \left(A - B\right)$

Subtracting $\frac{3}{2}$ from both ends of this second equation, we get:

$\frac{7}{2} = \frac{\sqrt{5}}{2} \left(A - B\right)$

Multiply both sides by $\frac{2}{\sqrt{5}}$ to get:

$\frac{7}{\sqrt{5}} = A - B$

Adding this to the first equation we find:

$2 A = 3 + \frac{7}{\sqrt{5}} = 3 + \frac{7}{5} \sqrt{5} \text{ }$ so $\text{ } A = \frac{3}{2} + \frac{7}{10} \sqrt{5}$

Subtracting from the first equation we find:

$2 B = 3 - \frac{7}{\sqrt{5}} = 3 - \frac{7}{5} \sqrt{5} \text{ }$ so $\text{ } B = \frac{3}{2} - \frac{7}{10} \sqrt{5}$

Hence the general formula for the given sequence can be written:

${a}_{n} = \left(\frac{3}{2} + \frac{7}{10} \sqrt{5}\right) {\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{n - 1} + \left(\frac{3}{2} - \frac{7}{10} \sqrt{5}\right) {\left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right)}^{n - 1}$