Suppose that each malefemale pair of rabbits in a farm produ

Suppose that each male/female pair of rabbits in a farm produces two new male/female pairs of rabbits at the age of 1 month and six new male/female pairs of rabbits at the age of 2 months and every month afterward. Assume that there is only one male/female pair of rabbits at the beginning of a year. Further assume that no rabbits die in the farm. Let a_n be the number of male/female pair of rabbits in the farm at the end of month n. Determine a recurrence relation for a_n and clearly justify your answer. By solving the recurrence relation in a) find the number of male/female pairs of rabbits in the farm at the end of month n. Prove the result in b) by strong mathematical induction. Let h_n equal the number of different ways in which the squares of a 1-by-n chessboard can be colored, using the colours red, white, and blue so that no two squares that are colored red are adjacent. Assume that h_0 = 1. Determine a recurrence relation for h_n and clearly justify your answer. Solving the recurrence relation in (a). Prove the result in (b) by strong mathematical induction.

Solution

8 a) The relation is a_n=2a_n-1+2

b) at end of month n, no fo male/female rabbits=2+2²+2³+....+2ⁿ

c) The below is the progression of male/female rabbits

1 month - 2, 2 month - 4+2= 2²+2, 3 months = 2³+2²+2, 4 months = 2⁴+2³+2²+2

So in n months = 2+2²+2³+....+2ⁿ

9 a) The set of colorings - red, white and blue can be divided into two categories. Those that end in a blue or white square start with coloring of length n1. Those that end in a red square start with coloring of length n2. So it can be colored like this - a white or blue square, then the red square.

This gives the recurrence relation h_n = 2h_n-1+2h_n-2.

b) so h0 = 1, h1 = 3, h2 = 8. The characteristic polynomial is r²2r2, which has roots r₁ = 1+ 3 and r₂= 1 3; hence the general solution is h_n = (r₁)ⁿ + (r₂)ⁿ.

c)Applying the initial conditions, 1 = h0 = + ,

3 = h1 = r + r2, which has solution = (3 + 2 3)/6, = (3 2 3)/6.

Hence the solution is h_n = (3+2 3)/6 (1 + 3)ⁿ+ (32 3)/6 (1 3)ⁿ .