319 Example Find the sum

1 ·2 + 2 ·3 + 3 ·4 + ··· + 99 ·100.

Solution: Observe that

k(k + 1) = 13 (k)(k + 1)(k + 2) − 13 (k −1)(k)(k + 1).

Therefore

7.2 First Order Recursions

The order of the recurrence is the difference between the highest and the lowest subscripts. For example un+2 −un+1 = 2

is of the first order, and

un+4 + 9u2n = n5

is of the fourth order.

A recurrence is linear if the subscripted letters appear only to the first power. For example

un+2 −un+1 = 2

is a linear recurrence and

x2n + nxn−1 = 1 and xn + 2xn−1 = 3

are not linear recurrences.

A recursion is homogeneous if all its terms contain the subscripted variable to the same power. Thus xm+3 + 8xm+2 −9xm = 0

is homogeneous. The equation

xm+3 + 8xm+2 −9xm = m2 −3

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is not homogeneous.

A closed form of a recurrence is a formula that permits us to find the n-th term of the recurrence without having to know a priori the terms preceding it.

We outline a method for solving first order linear recurrence relations of the form

320 Example Let x0 = 7 and xn = 2xn−1, n ≥ 1. Find a closed form for xn.

Solution: Raising subscripts we have the characteristic equation xn = 2xn−1. Cancelling, x = 2. Thus we try a solution of the form xn = A2n, were A is a constant. But 7 = x0 = A20 and so A = 7. The solution is thus xn = 7(2)n.

Aliter: We have

x0 = 7

Multiplying both columns,

x0x1 ···xn = 7 ·2n x0x1x2 ···xn−1.

Cancelling the common factors on both sides of the equality,

xn = 7 ·2n.

321 Example Let x0 = 7 and xn = 2xn−1 + 1, n ≥ 1. Find a closed form for xn.

Solution: By raising the subscripts in the homogeneous equation we obtain xn = 2xn−1 or x = 2. A solution to the homogeneous equation

will be of the form xn = A(2)n. Now f (n) = 1 is a polynomial of degree 0 (a constant) and so we test a particular constant solution C. The general solution will have the form xn = A2n + B. Now, 7 = x0 = A20 + B = A + B. Also, x1 = 2x0 + 7 = 15 and so 15 = x1 = 2A + B. Solving the simultaneous equations

A + B = 7,

2A + B = 15, we find A = 8, B = −1. So the solution is xn = 8(2n) −1 = 2n+3 −1.

Aliter: We have:

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Adding both columns, cancelling, and adding the geometric sum,

xn = 7 ·2n + (1 + 2 + 22 + ···+ 2n−1) = 7 ·2n + 2n −1 = 2n+3 −1.

Aliter: Let un = xn + 1 = 2xn−1 + 2 = 2(xn−1 + 1) = 2un−1. We solve the recursion un = 2un−1 as we did on our first example: un = 2nu0 = 2n(x0 + 1) = 2n ·8 = 2n+3. Finally, xn = un −1 = 2n+3 −1.

322 Example Let x0 = 2, xn = 9xn−1 −56n + 63. Find a closed form for this recursion.

Solution: By raising the subscripts in the homogeneous equation we obtain the characteristic equation xn = 9xn−1 or x = 9. A solution to the homogeneous equation will be of the form xn = A(9)n. Now f (n) = −56n + 63 is a polynomial of degree 1 and so we test a particular solution of the form Bn + C. The general solution will have the form xn = A9n + Bn + C. Now

x0 = 2, x1 = 9(2) −56 + 63 = 25, x2 = 9(25) −56(2) + 63 = 176. We thus solve the system

2 = A + C,

25 = 9A + B + C,

176 = 81A + 2B + C. We find A = 2, B = 7,C = 0. The general solution is xn = 2(9n) + 7n.

323 Example Let x0 = 1, xn = 3xn−1 −2n2 + 6n −3. Find a closed form for this recursion.

Solution: By raising the subscripts in the homogeneous equation we obtain the characteristic equation xn = 3xn−1 or x = 9. A solution to the homogeneous equation will be of the form xn = A(3)n. Now f (n) = −2n2 + 6n −3 is a polynomial of degree 2 and so we test a particular

solution of the form Bn2 + Cn + D. The general solution will have the form xn = A3n + Bn2 + Cn + D. Now

x0 = 1, x1 = 3(1) −2 + 6 −3 = 4, x2 = 3(4) −2(2)2 + 6(2) −3 = 13, x3 = 3(13) −2(3)2 + 6(3) −3 = 36. We thus solve the system

1 = A + D,

4 = 3A + B + C + D,

13 = 9A + 4B + 2C + D,

36 = 27A + 9B + 3C + D.

We find A = B = 1,C = D = 0. The general solution is xn = 3n + n2.

324 Example Find a closed form for xn = 2xn−1 + 3n−1, x0 = 2.

Solution: We test a solution of the form xn = A2n + B3n. Then x0 = 2, x1 = 2(2) + 30 = 5. We solve the system

2 = A + B,

7 = 2A + 3B.

We find A = 1, B = 1. The general solution is xn = 2n + 3n.

We now tackle the case when a = 1. In this case, we simply consider a polynomial g of degree 1 higher than the degree of f .

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325 Example Let x0 = 7 and xn = xn−1 + n, n ≥ 1. Find a closed formula for xn.

Solution: By raising the subscripts in the homogeneous equation we obtain the characteristic equation xn = xn−1 or x = 1. A solution to the homogeneous equation will be of the form xn = A(1)n = A, a constant. Now f (n) = n is a polynomial of degree 1 and so we test a particular solution of the form Bn2 + Cn + D, one more degree than that of f . The general solution will have the form xn = A + Bn2 + Cn + D. Since A and D are constants, we may combine them to obtain xn = Bn2 + Cn + E. Now, x0 = 7, x1 = 7 + 1 = 8, x2 = 8 + 2 = 10. So we solve the system

xn = 7 + n(n + 1) .

2

Some non-linear first order recursions maybe reduced to a lin ear first order recursion by a suitable transformation.

327 Example (Putnam 1985) Let d be a real number. For each integer m ≥ 0, define a sequence am ( j), j = 0, 1, 2, ··· by am (0) = 2dm , and am( j + 1) = (am( j + 1))2 + 2am( j), j ≥ 0. Evaluate

lim an(n). n→∞

Solution: Observe that am( j + 1) + 1 = (am( j))2 + 2am ( j) + 1 = (am( j) + 1)2. Put v j = am( j) + 1. Then v j+1 = v2j , and ln v j+1 = 2 ln v j ;

Put y j = ln v j. Then y j+1 = 2y j ; and hence 2ny0 = yn or 2n ln v0 = ln vn or vn = (v0)2n = (1 + 2dm )2n or am(n) + 1 = (1 + 2dm )2n . Thus an(n) = ( 2dn + 1)2n −1 → ed −1 as n → ∞.

7.3 Second Order Recursions

All the recursions that we have so far examined are first order recursions, that is, we find the next term of the sequence give n the preceding one. Let us now briefly examine how to solve some second order r ecursions.

We now outline a method for solving second order homogeneous linear recurrence relations of the form

xn = axn−1 + bxn−2 .

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