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After 4 months, the second rabbit also delivers a new rabbit, so two new rabbits are added. This means that the farmer will have 5 rabbits during the ﬁfth month. , who were already there after during the (k −1)-st month. In other words, if we denote by Fn the number of rabbits during the n-th month, then we have, for n = 2, 3, 4, . , Fn+1 = Fn + Fn−1 . (14) We also know that F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5. It is convenient to deﬁne F0 = 0; then equation (14) will remain valid for n = 1 as well.

Nk ! 15 We can describe the procedure of distributing the presents as follows. First, we n ways. select n1 presents and give them to the ﬁrst child. This can be done in n1 Then we select n2 presents from the remaining n − n1 and give them to the second child, etc. Complete this argument and show that it leads to the same result as the previous one. 16 The following special cases should be familiar from previous problems and theorems. Explain why. (a) n = k, n1 = n2 = . . = nk ; (b) n1 = n2 = .

B) F0 − F1 + F2 − F3 + . . − F2n−1 + F2n = F2n−1 − 1. (c) F02 + F12 + F22 + . . + Fn2 = Fn · Fn+1 . (d) Fn−1 Fn+1 − Fn2 = (−1)n . 6 Mark the ﬁrst entry (a 1) of any row of the Pascal triangle. Move one step East and one step Northeast, and mark the entry there. Repeat this until you get out of the triangle. Compute the sum of the entries you marked. (a) What numbers do you get if you start from diﬀerent rows? First ”conjecture”, than prove your answer. (b) Formulate this fact as an identity involving binomial coeﬃcients.

### Discrete Mathematics by Yale

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