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)