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)#