1.Find the characteristic equation by “raising the subscri pts” in the form xn = axn−1 + bxn−2. Cancelling this gives x2 −ax −b = 0. This equation has two roots r1 and r2.

1 = A,

4 = 2A + 2B.

This solves to A = 1, B = 1. The solution is thus xn = 2n + n2n .

7.4 Applications of Recursions

331 Example Find the recurrence relation for the number of n digit binary sequences with no pair of consecutive 1's.

Solution: It is quite easy to see that a1 = 2, a2 = 3. To form an, n ≥ 3, we condition on the last digit. If it is 0, the number of sequences sought is an−1. If it is 1, the penultimate digit must be 0, and the number of sequences sought is an−2. Thus

an = an−1 + an−2, a1 = 2, a2 = 3.

332 Example Let there be drawn n ovals on the plane. If an oval intersects each of the other ovals at exactly two points and no three ovals intersect at the same point, find a recurrence relation for th e number of regions into which the plane is divided.

Solution: Let this number be an. Plainly a1 = 2. After the n −1th stage, the nth oval intersects the previous ovals at 2(n −1) points, i.e. the nth oval is divided into 2(n −1) arcs. This adds 2(n −1) regions to the an−1 previously existing. Thus

an = an−1 + 2(n −1), a1 = 2.

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333 Example Find a recurrence relation for the number of regions into which the plane is divided by n straight lines if every pair of lines intersect, but no three lines intersect.

Solution: Let an be this number. Clearly a1 = 2. The nth line is cut by he previous n −1 lines at n −1 points, adding n new regions to the previously existing an−1. Hence

an = an−1 + n, a1 = 2.

334 Example (Derangements) An absent-minded secretary is filling n envelopes with n letters. Find a recursion for the number Dn of ways in which she never stuffs the right letter into the right envelope.

Solution: Number the envelopes 1, 2, 3, ··· , n. We condition on the last envelope. Two events might happen. Either n and r(1 ≤ r ≤ n −1) trade places or they do not.

In the first case, the two letters r and n are misplaced. Our task is just to misplace the other n −2 letters, (1, 2, ··· , r −1, r + 1, ··· , n −1) in the slots (1, 2, ··· , r −1, r + 1, ··· , n −1). This can be done in Dn−2 ways. Since r can be chosen in n −1 ways, the first case can happen in

ways.

In the second case, let us say that letter r, (1 ≤ r ≤ n −1) moves to the n-th position but n moves not to the r-th position. Since r has been misplaced, we can just ignore it. Since n is not going to the r-th position, we may relabel n as r. We now have n −1 numbers to misplace, and this can be done in Dn−1 ways.

As r can be chosen in n −1 ways, the total number of ways for the second case is (n −1)Dn−1 . Thus Dn = (n −1)Dn−2 + (n −1)Dn−1 .

335 Example There are two urns, one is full of water and the other is empty. On the first stage, half of the contains of urn I is passed into urn II. On the second stage 1/3 of the contains of urn II is passed into urn I. On stage three, 1/4 of the contains of urn I is passed into urn II. On stage four 1/5 of the contains of urn II is passed into urn I, and so on. What fraction of water remains in urn I after the 1978th stage?

Solution: Let xn, yn, n = 0, 1, 2, . . . denote the fraction of water in urns I and II respectively at stage n. Observe that xn + yn = 1 and that

x0 = 1; y0 = 0

x1 = x0 − 12 x0 = 12 ; y1 = y1 + 12 x0 = 12 x2 = x1 + 13 y1 = 23 ; y2 = y1 − 13 y1 = 13 x3 = x2 − 14 x2 = 12 ; y1 = y1 + 14 x2 = 12 x4 = x3 + 15 y3 = 35 ; y1 = y1 − 15 y3 = 25 x5 = x4 − 16 x4 = 12 ; y1 = y1 + 16 x4 = 12 x6 = x5 + 17 y5 = 47 ; y1 = y1 − 17 y5 = 37 x7 = x6 − 18 x6 = 12 ; y1 = y1 + 18 x6 = 12 x8 = x7 + 19 y7 = 59 ; y1 = y1 − 19 y7 = 49

A pattern emerges (which may be proved by induction) that at each odd stage n we have xn = yn = 12 and that at each even stage we have (if

Homework

336 Problem Find the sum of all the integers from 1 to 1000 inclusive, which are not multiples of 3 or 5.

337 Problem The sum of a certain number of consecutive positive integers is 1000. Find these integers. (There is more than one solution. You must find them all.)

338 Problem Use the identity

n5 −(n −1)5 = 5n4 −10n3 + 10n2 −5n + 1.

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Answers

336 We compute the sum of all integers from 1 to 1000 and weed out the sum of the multiples of 3 and the sum of the multiples of 5, but put back the multiples of 15, which we have counted twice. Put

An = 1 + 2 + 3 + ··· + n,

B = 3 + 6 + 9 + ··· + 999 = 3A333,

C= 5 + 10 + 15 + ··· + 1000 = 5A200,

D= 15 + 30 + 45 + ··· + 990 = 15A66.

The desired sum is

opposite parity. Since 2000 = 2 5 , the odd factors of 2000 are 1, 5, 25, and 125. We then see that the problem has the following solutions: n = 1, l = 999,

n = 5, l = 197, n = 16, l = 54, n = 25, l = 27.

338 Using the identity for n = 1 to n:

n5 = 5s4 −10s3 + 10s2 −5s1 + n,

whence

=n5 + n2(n + 1)2 − n(n + 1)(2n + 1) + n(n + 1) − n

=n5 + n4 + n3 − n .

5 2 3 30

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