Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Discrete Mathematics
October 16, 2015
Quiz 8
1. Show (by induction) the following equation
n
X
i=1
1
i(i + 1) = n(n + 1)(n + 2)
3
for each positive integer n.
Proof. Basis step: The case n = 1 is true is true because both
sides of the equation
1
X
i(i + 1) = 1(1 + 1) = 2 and
1
3
· 1 · (1 + 1)(1 + 2) = 31 6 = 2
i=1
evaluate to 2.
Induction hypothesis: Suppose the result is true for k ∈ N;
that is, we assume we have
k
X
i=1
1
i(i + 1) = 1(1 + 1) + 2(2 + 1) + · · · + k(k + 1) = k(k + 1)(k + 2)
3
We must prove
k+1
X
1
(k + 1)(k + 1 + 1)(k + 1 + 2)
3
1
=
(k + 1)(k + 2)(k + 3)
3
i(i + 1) =
i=1
We have
k+1
X
i(i + 1) = 1(1 + 1) + 2(2 + 1) + · · · + k(k + 1) +(k + 1)(k + 1 + 1)
|
{z
}
i=1
Pk
i=1
i(i+1)= 31 k(k+1)(k+2)
1
k(k + 1)(k + 2) + (k + 1)(k + 2)
3
1
1
3
=
k + 1 (k + 1)(k + 2) =
k+
(k + 1)(k + 2)
3
3
3
=
=
1
1
(k + 3)(k + 1)(k + 2) = (k + 1)(k + 2)(k + 3).
3
3
This is what we had to show.
2. Consider the sequence defined recursively by
a0 = 1 and, for n > 0, let an = 3an−1 − 1.
Prove: an =
3n + 1
.
2
Proof. Basis step: The case n = 1 is true is true because both
sides of the equation
a0 = 1 and
2
30 + 1
= =1
2
2
are equal.
Induction hypothesis: Suppose the result is true for k ∈ N;
that is, we assume we have
ak =
We must prove ak+1 =
3k + 1
.
2
3k+1 + 1
2
We have
ak+1 = 3ak+1−1 − 1 = 3ak − 1 (by the definition of ak+1 )
k
3 +1
3k + 1
)
= 3
− 1 (by the assumption ak =
2
2
3 · 3k + 3
=
−1
2
·3k+1 + 3 2
−
=
2
2
·3k+1 + 3 − 2
=
2
·3k+1 + 1
=
2
This is what we had to show.
2. Consider the sequence defined recursively by
a1 = 1 , a2 = 8 and, for n ≥ 3, let an = an−1 + 2an−2 .
(a) What are the first five terms of the sequence
a1 = 1
a2 = 8
a3 = 10
a4 = 26
a5 = 46
(b) Using strong induction prove an = 3 · 2n−1 + 2(−1)n .
Proof. Basis case:
• When n = 1 then a1 = 3 · 21−1 + 2(−1)1 = 3 · 20 + 2(−1)1 =
3 − 2 = 1, as required.
• When n = 2 then a2 = 3 · 22−1 + 2(−1)2 = 3 · 21 + 2 =
6 + 2 = 8, as required.
Induction hypothesis (strong): Suppose the formula holds
for values k − 1 and k, i.e. it holds
• ak−1 = 3 · 2k−1−1 + 2(−1)k−1 = 3 · 2k−2 + 2(−1)k−1
• ak = 3 · 2k−1 + 2(−1)k
and
To prove: ak+1 = 3 · 2k+1−1 + 2(−1)k+1 = 3 · 2k + 2(−1)k+1 .
We calculate,
ak+1 = ak+1−1 + 2ak+1−2 = ak + 2ak−1 (by the definition of ak+1 )
k−1
k
k−2
k−1
= 3·2
+ 2(−1) + 2 3 · 2
+ 2(−1)
(by induction hypothesis )
= 3 · 2k−1 + 2 (−1)k +3 · 2 · 2k−2 + 4(−1)k−1
| {z }
k−1
= 3·2
(−1)(−1)k−1
k−1
+ 2(−1)(−1)
+ 3 · 2k−2+1 + 4(−1)k−1
= 3 · 2k−1 − 2(−1)k−1 + 3 · 2k−1 + 4(−1)k−1
= |{z}
6 ·2k−1 + 2(−1)k−1 = 3 · 2 · 2k−1 + 2(−1)k−1 · (−1)2
| {z }
3·2
= 3 · 2k + 2(−1)k+1
as required.
1