CSci/Math2112
Assignment 7
Due July 10, 2015
Prove the following statements using induction.
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1. (BoP 10 #6) For every natural number n, it follows that
n
X
(8i − 5) = 4n2 − n.
i=1
Solution:
Proof. Proof (by induction)
n
X
Base case: If n = 1, then
(8i − 5) = 8 · 1 − 5 = 3 = 4 · 1 − 1 = 4n2 − n.
i=1
Induction step: Assume that
k
X
(8i − 5) = 4k 2 − k for some k ∈ N. Then
i=1
k+1
X
(8i − 5) =
i=1
!
k
X
(8i − 5) + 8(k + 1) − 5
i=1
= 4k 2 − k + 8k + 3
= 4k 2 + 7k + 3
= 4(k + 1)2 − (k + 1).
Thus by induction
n
X
(8i − 5) = 4n2 − n for all natural numbers n.
i=1
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2. (BoP 10 #10) For any integer n ≥ 0, it follows that 3 | 52n − 1 .
Solution:
Proof. Proof (by induction)
Base case: If n = 0, then 52n − 1 = 1 − 1 = 0 which is divisible by 3.
Induction step: Assume 3 | 52k − 1 for some integer k ≥ 0. Then there exists an integer a such that
52k − 1 = 3a.
We have
52(k+1) − 1 = 25 · 52k − 1
= 25 52k − 1 + 24
= 25(3a) + 24
= 3(25a + 8).
Since 25a + 8 ∈ Z, this shows that 3 | 52(k+1) − 1 .
Thus by induction we have 3 | 52n − 1 for all nonnegative integers n.
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2
3. (BoP 10 #20) We know that (1 + 2 + 3 + . . . + n) = 13 + 23 + 33 + . . . + n3 for every n ∈ N. (You may
n(n + 1)
assume that 1 + 2 + . . . + n =
for all n ∈ N.)
2
Solution:
Proof. Proof (by induction)
2
Base case: If n = 1, then (1 + 2 + 3 + . . . + n) = 12 = 1 = 13 = 13 + 23 + 33 + . . . + n3 .
2
Induction step: Assume that for some k ∈ N we know that (1 + 2 + 3 + . . . + k) = 13 +23 +33 +. . .+k 3 .
Then
2
2
(1 + 2 + 3 + . . . + k + (k + 1)) = (1 + 2 + 3 + . . . + k) + 2 (1 + 2 + 3 + . . . + k) (k + 1) + (k + 1)2
k(k + 1)
(k + 1) + (k + 1)2
2
= 13 + 23 + 33 + . . . + k 3 + k(k + 1)2 + (k + 1)2
= 13 + 23 + 33 + . . . + k 3 + 2
= 13 + 23 + 33 + . . . + k 3 + (k + 1)(k + 1)2
= 13 + 23 + 33 + . . . + k 3 + (k + 1)3 .
2
Thus by induction we know that (1 + 2 + 3 + . . . + n) = 13 + 23 + 33 + . . . + n3 for every n ∈ N
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4. The sequence a0 , a1 , a2 , . . . is defined recursively by a0 = 0 and an = 2an−1 + n − 1 for n ∈ N. Then
an = 2n − n − 1 for all non-negative integers n.
Solution:
Proof. Proof (by induction)
Base case: If n = 0, then a0 = 0 and 20 − 0 − 1 = 1 − 1 = 0. Thus an = 2n − n − 1 for n = 0.
Induction step: Assume that ak−1 = 2k−1 − (k − 1) − 1 for some k ≥ 1. Then
ak = 2ak−1 + k − 1
= 2 2k−1 − (k − 1) − 1 + k − 1
= 2k − 2k + k − 1
= 2k − k − 1.
Thus by induction we have an = 2n − n − 1 for all non-negative integers n.