Part 2

For 3 players or less, there is no lag time. So we consider for 4 players and above.



U4 = 3 = 4 - 1 = 4 - 1½ + ½ = 4 - 1½ + ½(1)
= 4 - 1½ + (½)(-1)⁴

U5 = 3 = 5 - 2 = 5 - 1½ - ½ = 5 - 1½ + ½(-1)
= 5 - 1½ + ½(-1)^5

U6 = 5 = 6 - 1 = 6 - 1½ + ½ = 6 - 1½ + ½(1)
= 6 - 1½ + ½(-1)^6

U7 = 5 = 7 - 2 = 7 - 1½ - ½ = 7 - 1½ + ½ (-1)
= 7 - 1½ + ½(-1)^7


This holds for the rest of the terms. Notice that when the term number is odd, there's a -½ and when it's even, there's a +½. The ½ oscillates between positive and negative.

(-1)ⁿ is negative when n is odd and is positive when n is even. We can use this for the general term to describe the ½ that is oscillating



So,

Un = n - 1½ + ½(-1)ⁿ
= ½(2n - 3 + (-1)ⁿ) for n ≥ 4


Part 3


Since Un = ½(2n - 3 + (-1)ⁿ),

Then U(n - 1)

= ½(2(n - 1) - 3 + (-1)ⁿ‾¹ )

= ½(2n - 5 + (-1)ⁿ‾¹ )

= ½(2n - 5 + (-1)‾¹ x (-1)ⁿ)

= ½(2n - 5 - (-1)ⁿ )

= ½( 4n - 8 - 2n + 3 - (-1)ⁿ )

= ½(4n - 8) - ½( 2n - 3 + (-1)ⁿ )

= 2n - 4 - Un

= 2(n - 2) - Un


So ,Un = 2(n - 2) - U(n-1) for n ≥ 5

Or U(n+1) = 2(n - 1) - Un for n ≥ 4





Alternatively,


U(n - 1) = ½(2n - 5 - (-1)ⁿ )

(From above)


The (-1)ⁿ can be eliminated by adding Un and U(n - 1) together.


Un + U(n - 1)

= ½(2n - 3 + (-1)ⁿ ) + ½(2n - 5 - (-1)ⁿ)

= ½(2n - 3 + (-1)ⁿ + 2n - 5 - (-1)ⁿ )

= ½(4n - 8)

= 2n - 4

= 2(n - 2)


So, Un = 2(n - 2) - U(n-1) for n ≥ 5

Or U(n+1) = 2(n - 1) - Un for n ≥ 4