62 Chapter 4
392 Problem
Find a closed form for x = 3
, x
n
= x
n−1
+ 4 3
.
393 Problem Find a closed form for x
= 1 , x
n
= 5x
n−1
− 20n + 25 .
394 Problem Find a closed form for x
= 1 , x
n
= x
n−1
+ 12n .
395 Problem Find a closed form for x
n
= 2x
n−1
+ 95
n−1
, x = 5
.
396 Problem Find a closed form for
a = 5
, a
j+1
= a
2 j
+ 2a
j
, j ≥ 0.
397 Problem AIME, 1994 If n
≥ 1, x
n
+ x
n−1
= n
2
. Given that x
19
= 94 , find the remainder when x
94
is divided by 1000 .
398 Problem
Find a closed form for x
= −1; x
n
= x
n−1
+ n
2
, n 0.
399 Problem If u
= 1 3 and u
n+1
= 2u
2 n
− 1 , find a closed form for u
n
.
400 Problem Let x
1
= 1 , x
n+1
= x
2 n
− x
n
+ 1 , n 0. Shew that
∞
X
n=1
1 x
n
= 1 .
4.6 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 given the preceding one. Let us now briefly examine how to solve some second order recursions.
We now outline a method for solving second order homogeneous linear recurrence relations of the form
x
n
= ax
n−1
+ bx
n−2
. 1. Find the characteristic equation by “raising the subscripts” in the form x
n
= ax
n−1
+ bx
n−2
. Cancelling this gives x
2
− ax − b = 0 . This equation has two roots r
1
and r
2
. 2. If the roots are different, the solution will be of the form x
n
= Ar
1 n
+ Br
2 n
, where A , B are constants.
3. If the roots are identical, the solution will be of the form x
n
= Ar
1 n
+ Bnr
1 n
.
401 Example Let x
= 1 , x
1
= −1 , x
n+2
+ 5x
n+1
+ 6x
n
= 0. Solution: The characteristic equation is x
2
+ 5x + 6 = x + 3x + 2 = 0. Thus we test a solution of the form x
n
= A−2
n
+ B−3
n
. Since 1 = x = A + B
, −1 = −2A − 3B, we quickly find A = 2, B = −1. Thus the solution is x
n
= 2−2
n
− −3
n
.
402 Example
Find a closed form for the Fibonacci recursion f = 0
, f
1
= 1 , f
n
= f
n−1
+ f
n−2
. Solution: The characteristic equation is f
2
− f − 1 = 0, whence a solution will have the form f
n
= A
1 + √
5 2
n
+ B
1 − √
5 2
n
. The initial conditions give 0 = A + B
, 1 = A
1 + √
5 2
+ B
1 − √
5 2
= 1
2 A + B +
√ 5
2 A − B =
√ 5
2 A − B
This gives A = 1
√ 5
, B = − 1
√ 5
. We thus have the Cauchy-Binet Formula: f
n
= 1
√ 5
1 + √
5 2
n
− 1
√ 5
1 − √
5 2
n
4.7
403 Example
Solve the recursion x = 1
, x
1
= 4 , x
n
= 4x
n−1
− 4x
n−2
= 0 .
Second Order Recursions 63
Solution: The characteristic equation is x
2
− 4x + 4 = x − 2
2
= 0. There is a multiple root and so we must test a solution of the form x
n
= A2
n
+ Bn2
n
. The initial conditions give 1 = A
, 4 = 2A + 2B
. This solves to A = 1
, B = 1. The solution is thus x
n
= 2
n
+ n2
n
.
64 Chapter 4
Practice
404 Problem Solve the recursion x
= 0 , x
1
= 1 , x
n
= 10x
n−1
− 21x
n−2
.
405 Problem Solve the recursion x
= 0 , x
1
= 1 , x
n
= 10x
n−1
− 25x
n−2
.
406 Problem Solve the recursion x
= 0 , x
1
= 1 , x
n
= 10x
n−1
− 21x
n−2
+ n.
407 Problem Solve the recursion x
= 0 , x
1
= 1 , x
n
= 10x
n−1
− 21x
n−2
+ 2
n
.
4.7 Applications of Recursions
