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

##### 2 Answers

nth term plus the nth + 1 term:

#### Explanation:

This sequence is the:

nth term plus the nth + 1 term:

This is also called the Fibonacci Series.

The general term is given by the formula:

#a_n = (3/2+7/10sqrt(5))(1/2+sqrt(5)/2)^(n-1) + (3/2-7/10sqrt(5))(1/2-sqrt(5)/2)^(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, color(blue)(3, 5, 8, 13, 21), 34, 55, 89, 144,...#

Note that the given sequence starts at

#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,...#

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

#x^2 = 1 + x#

So:

#0 = x^2-x-1 = (x-1/2)^2-(sqrt(5)/2)^2 = (x-1/2-sqrt(5)/2)(x-1/2+sqrt(5)/2)#

Hence:

Consider the sequence:

#b_n = A(1/2+sqrt(5)/2)^(n-1) + B(1/2-sqrt(5)/2)^(n-1)#

where

Notice that any such sequence

#b_(n+2) = b_n + b_(n+1)#

So if we can find values for

So we just require:

#3 = b_1 = A(1/2+sqrt(5)/2)^(1-1) + B(1/2-sqrt(5)/2)^(1-1)#

#color(white)(3) = A+B#

#5 = b_2 = A(1/2+sqrt(5)/2)^(2-1) + B(1/2-sqrt(5)/2)^(2-1)#

#color(white)(5) = (1/2+sqrt(5)/2)A + (1/2-sqrt(5)/2)B#

#color(white)(5) = 1/2(A+B) + sqrt(5)/2(A-B)#

#color(white)(5) = 3/2 + sqrt(5)/2(A-B)#

Subtracting

#7/2 = sqrt(5)/2(A-B)#

Multiply both sides by

#7/sqrt(5) = A-B#

Adding this to the first equation we find:

#2A = 3+7/sqrt(5) = 3+7/5sqrt(5)" "# so#" "A = 3/2+7/10sqrt(5)#

Subtracting from the first equation we find:

#2B = 3-7/sqrt(5) = 3-7/5sqrt(5)" "# so#" "B = 3/2-7/10sqrt(5)#

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

#a_n = (3/2+7/10sqrt(5))(1/2+sqrt(5)/2)^(n-1) + (3/2-7/10sqrt(5))(1/2-sqrt(5)/2)^(n-1)#