In the first sixteen terms of the Fibonacci sequence, which ones are even numbers?
1 1 2 3 5 8 13 21 34 55 ...
notice every third number is even
can you figure it out from there?
If not, just list the rest of the first 16 and count them.
Thank You!
To determine which terms of the Fibonacci sequence are even numbers, we first need to understand what the Fibonacci sequence is. The Fibonacci sequence is a sequence of numbers in which each number is the sum of the two preceding ones, usually starting with 0 and 1.
To find the first sixteen terms of the Fibonacci sequence, we can apply the formula:
F(n) = F(n-1) + F(n-2)
where F(1) = 0 and F(2) = 1.
We can start by manually calculating the first few terms of the sequence:
F(1) = 0
F(2) = 1
F(3) = F(2) + F(1) = 1 + 0 = 1
F(4) = F(3) + F(2) = 1 + 1 = 2
F(5) = F(4) + F(3) = 2 + 1 = 3
Continuing this pattern, we can calculate the first sixteen terms of the Fibonacci sequence:
F(6) = F(5) + F(4) = 3 + 2 = 5
F(7) = F(6) + F(5) = 5 + 3 = 8
F(8) = F(7) + F(6) = 8 + 5 = 13
F(9) = F(8) + F(7) = 13 + 8 = 21
F(10) = F(9) + F(8) = 21 + 13 = 34
F(11) = F(10) + F(9) = 34 + 21 = 55
F(12) = F(11) + F(10) = 55 + 34 = 89
F(13) = F(12) + F(11) = 89 + 55 = 144
F(14) = F(13) + F(12) = 144 + 89 = 233
F(15) = F(14) + F(13) = 233 + 144 = 377
F(16) = F(15) + F(14) = 377 + 233 = 610
Now, we can determine which terms are even numbers from the above list. An even number is divisible by 2, meaning it leaves no remainder when divided by 2.
Looking at the list, we can see that the even numbers among the first sixteen terms are:
F(2), F(4), F(6), F(8), F(10), F(12), F(14), and F(16).
So, the even numbers in the first sixteen terms of the Fibonacci sequence are: 0, 2, 8, 34, 144, 610.