Question

Suppose you have the following recursion formula a1 = 1, a2 = 2, and an = a(n – 1)+ a(n – 2)for integers n ≥ 3. How would you determine the next three terms?

Answer 1

follow the rules given by the equations

Answer 2

ok

Answer 3

I'm new to this

Answer 4

notice that the value of n determines how the equation looks

an = a(n – 1)+ a(n – 2)

we already know the first 2 values for n=1 and n=2; a1 and a2 are given

Answer 5

when n=3, what does the setup look like?

Answer 6

3a=1 4a=-1

Answer 7

a3 = a(3 – 1)+ a(3 – 2)
= a2 +a1
= (2) +(1)
= 3

Answer 8

so, a1 = 1; a2 = 2; a3 = 3

run the rule again, using n=4

a4 = a(4 – 1)+ a(4 – 2)
= a3 + a2
= 3 +2
= 5

Answer 9

oh ok

Answer 10

can you determine a5 for me? when n=5?

Answer 11

is it 45
\

Answer 12

show me your process, not just your end results. the process is what is important ... and no it aint 45

Answer 13

ok this is what I have
a1 = a(3-1)+a(3-2)--> 1a(2)+1a(1)
a2 = a(3-1)+a(3-2)--> 2(2)+2(1)= 4+2 = 6 which will become a3 set
a3 which is #6 = a(3-1)+a(3-2)--> 6(2)+6(1)= 12+6 = 18 will become a4 set.
a4 which is #18 = a(3-1)+a(3-2)--> 18(2)+ 18(1) = 54 which will become a5 set.

Answer 14

i see what you did, your confusing the notation for multiplication i believe

Answer 15

yes thts where I got lost

Answer 16

we have a sequence A defined by the elements: \({a_1,a_2,a_3,...,a_n}\), the subscripts just refer to the 1st, or 2nd, or .... nth, term in the set.

the rule given to generate the set is:\[a_n=a_{n-1}+a_{n-2}\]such that we already know the first 2 elements are: \(a_1=1,~a_2=2\).

Answer 17

notice that the rule is just stating in math terms that the next value is the sum of the 2 values that came before it

Answer 18

ok

Answer 19

so \(a_3\) is equal to the sum of \(a_2\) and \(a_1\).

\[a_3=1+3 = 3\]\[a_4=a_3+a_2\]\[a_5=a_4+a_3\]\[a_6=a_5+a_4\]
etc ...

Answer 20

thank you

Answer 21

u made it seem easy

Answer 22

:) sometimes math is like learning a whole new language

Answer 23

yes very much so