Formal Methods
Homework Assignment 9, Part 3
April 11, 2007
41. Let x1 = 1, and for n > 1, let xn =
√
3xn−1 + 1. Prove that xn < 4 for all n ∈ N.
Proof. We use the basic PMI. The base case is clear: if n = 1, xn = 1, which is less
than 4. So xn < 4 when n = 1.
Suppose then that n is a positive integer such that xn < 4. We need to see that
xn+1 < 4. Since n ≥ 1, n + 1 > 1. So, by definition,
p
xn+1 =
3x(n+1)−1 + 1
√
3xn + 1
=
√
√
Now
√ the function y = 3x + 1 is an increasing function. So if a < b, then 3a + 1 ≤
3b + 1. Thus, since xn < 4,
√
3·4+1
xn+1 ≤
√
13
=
√
<
16
= 4
So by the PMI, xn < 4, for all positive integers n.
54. Let a1 = 1, and for each natural number n > 1, let an = 3an−1 − 1. Prove that for each
natural number n,
1 n−1
an =
3
+1 .
2
Proof. If n = 1, an = 1, and
1
1 n−1
3
+ 1 = · 2 = 1.
2
2
So if n = 1,
an =
1 n−1
3
+1 .
2
1
Suppose then that n is a positive integer such that
an =
We need to show that
an+1 =
1 n−1
3
+1 .
2
1 (n+1)−1
3
+1 .
2
Since n ≥ 1, n + 1 > 1. So
an+1 = 3a(n+1)−1 − 1
= 3an − 1
1 n−1
3
+1 −1
= 3
2
3
1
=
· 3 · 3n−1 + − 1
2
2
1 n 1
·3 +
=
2
2
1 (n+1)−1
=
3
+1
2
So by the PMI
an =
1 n−1
3
+1 ,
2
for all positive integers n.
55. Let a1 = 1 and a2 = 3, and for each natural number n > 2, let an = 3an−1 − 2an−2 .
Prove that for each natural number n, an = 2n − 1.
Proof. If n = 1, an = 1, and 2n − 1 = 2 − 1 = 1. So if n = 1, an = 2n − 1.
[In the induction step, we would ordinarily assume that n is a positive integer such that
an = 2n − 1, and then try to deduce that an+1 = 2n+1 − 1. However, this would involve
cases, because there are two possible formulas for an+1 : the formula if n + 1 = 2 (i.e.,
n = 1) and the formula if n + 1 > 2. To avoid this we could prove that a2 = 22 − 1
directly, and then, in the induction step, instead of assuming n ≥ 1, we would assume
n ≥ 2. Either approach is fine. I’ll use the second approach.]
If n = 2, an = 3, and 2n − 1 = 22 − 1 = 3. So in this case also, an = 2n − 1.
[Now if n ≥ 2, the formula for an+1 involves both an and an−1 . So instead of using the
basic PMI, I’ll use the strong PMI.]
Suppose then that n is an integer ≥ 2 such that ak = 2k + 1, whenever k = 1, 2, . . . , n.
Then, since n ≥ 2, n + 1 > 2. So
an+1 = 3a(n+1)−1 − 2a(n+1)−2
= 3an − 2an−1
= 3(2n − 1) − 2(2n−1 − 1)
2
=
=
=
=
3 · 2n − 3 − 2 · 2n−1 + 2
3 · 2n − 2n − 1
2 · 2n − 1
2n+1 − 1
So by the strong PMI, for each positive integer n, an = 2n − 1.
57. Let a1 = a2 = a3 = 1, and for each natural number n > 3, let
an = an−1 + an−2 + an−3 .
Prove that for each natural number n > 1, an ≤ 2n−2 .
Proof. By analogy with problem 55, we use the strong PMI and we prove three base
cases. If n = 2, an = 1 and 2n−2 = 1. If n = 3, an = 1, and 2n−2 = 2. If n = 4,
an = an−1 + an−2 + an−3 = a3 + a2 + a1 = 3,
and 2n−2 = 4. So if n = 2, 3, or 4, an ≤ 2n−2 .
Suppose then that n is an integer ≥ 4 such that ak ≤ 2k−2 for k = 2, 3, . . . n. We want
to see that an+1 ≤ 2(n+1)−2 . Since n ≥ 4, n + 1 > 5. So
an+1 = a(n+1)−1 + a(n+1)−2 + a(n+1)−3
= an + an−1 + an−2
Since n ≥ 4, n, n − 1, and n − 2 are all ≥ 2. So the induction hypothesis applies to an ,
an−1 , and an−2 , and
an+1 =
≤
≤
=
=
=
=
an + an−1 + an−2
2n−2 + 2n−3 + 2n−4
2n−2 + 2n−3 + 2 · 2n−4
2n−2 + 2 · 2n−3
2n−2 + 2n−2
2 · 2n−2
2(n+1)−2
So by the strong PMI, an ≤ 2n−2 for all integers n ≥ 2.
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