Fibonacci posed the following problem: suppose that rabbits live forever and that every month each pair produces a new pair which becomes productive at age 2 months. if we start with one newborn pair, how many pairs of rabbits will we have in the nth month?

There will be as many pairs as the value of the nth term of the Fibonacci sequence

Step-by-step explanation:

During a month, only those pairs of rabbits that were given birth the previous month dont produce their pair. For the month n you take the amount of pairs of the n-1 month (lets call it an-1) and we have to add the new pairs created, that were created for rabits given birth on the month n-2 or before (in other words, the active ones). This means that an-2 pairs were created, so the total number of pairs, an, is given by the formula

(n has to be greater than 2)

or, equivalently

This is Fibonacci's formula for values greater than 2, also note that

a1 = 1

a2 = 1 ( because the pair was inactive this month)

a3 = 2 (because now the pair is activa)

As a result, we have

a1 = 1

a2 = 1

an+2 = an+1 - an (with n at least 1)

Thus, there were as many pairs in the nth month as the value of the nth term in the Fibonacci sequence.

the solution is 12x-6y

35x 3= 105

step-by-step explanation:

35+35= 70   70 + 35= 105