Homework 3 solutions
MATH 2283 Spring 2015
4.6 (2 pts)
Let P (n) be the statement “52n − 1 is divisible by 8”.
P (1) is clearly true since 52(1) − 1 = 24 is divisible by 8.
Assume P (n) is true for some n.
Show P (n + 1) : 52(n+1) − 1 is divisible by 8.
52(n+1) − 1 = 52n+2 − 1 = 52 52n − 1 = 52 (52n − 1) +24
| {z }
divisible by 8
by ind. hyp.
Since 24 is also divisible by 8, their sum 52(n+1) − 1 is divisible by 8. Thus by
induction, 52n − 1 is divisible by 8 ∀n ∈ N.
4.10 (2 pts)
P (n) : 1 + 3 + ... + (2n − 1) = n2 .
P (1): (1)2 = 12 is true.
Assume P (n) is true for some n.
Show P (n + 1) is true:
P (n + 1) ⇐⇒
1 + 3 + ... + (2n − 1) +(2(n + 1) − 1) = (n + 1)2
|
{z
}
=n2 by induction hypothesis
⇐⇒ n2 + 2n + 1 = (n + 1)2
which is true.
Thus P (n + 1) is true, so by induction, P (n) holds true for all n ∈ N.
4.17 (b) (2 pts)
P (n) : 3n > n3 .
P (4) : 34 = 81 > 43 = 64 is true.
Assume P (n) is true for some n.
Show P (n + 1) is true:
P (n + 1) ⇐⇒ 3n+1 > (n + 1)3
⇐⇒ 3 · 3n > n3 + 3n2 + 3n + 1
⇐⇒ 2 · 3n > 3n2 + 3n + 1.
This is true since 2 · 3n > 2 · n3 by induction hypothesis and 2 · n3 > 3n2 + 3n + 1
by part (a). Thus by induction 3n > n3 ∀n ∈ N.
1
4.20 (b) (3 pts) We show that ∀n ∈ N, the statement P (n) : cn ∈ N & dn ∈ N
is true, where
√
√
√
cn = b[(a + b)n − (a − b)n ]
√
√
dn = (a + b)n + (a − b)n .
P (1) : c1 = 2b ∈ N & d1 = 2a ∈ N is true.
Assume P (n) is true for some n.
Show P (n + 1) : cn+1 ∈ N & dn+1 ∈ N is true:
√
√
√
cn+1 = b[(a + b)(n+1) − (a − b)(n+1) ]
√
√
√
√
√
= b[(a + b)(a + b)n − (a − b)(a − b)n ]
√
√
√
√
√
√
√
= b[a(a + b)n + b(a + b)n − a(a − b)n + b(a − b)n ]
= acn + bdn
∈N
by induction hypothesis
and similarly
√
√
b)(n+1) + (a − b)(n+1)
√
√
√
√
= (a + b)(a + b)n + (a − b)(a − b)n
√
√
√
√
√
√
= a(a + b)n + b(a + b)n + a(a − b)n − b(a − b)n
dn+1 = (a +
= adn + cn
∈N
by induction hypothesis.
Therefore by induction, cn , dn ∈ N for all n ∈ N.
4.22
(1 pt)