Graded Homework # 6 Solutions
Section 4.1
(3) (a) If n = 1, P (1) is the statement “12 = 1(1 + 1)(2 + 1)/6”.
(b) P (1) is true, since 12 = 1 and 1(1 + 1)(2 + 1)/6 = 1.
(c) The inductive hypothesis is the assumption that P (k) is true for some k ≥ 1. Here, P (k)
is the statement “12 + 22 + · · · + k 2 = k(k + 1)(2k + 1)/6”.
(d) For the inductive step, we must show
12 + 22 + · · · + k 2 + (k + 1)2 = (k + 1)((k + 1) + 1)(2(k + 1) + 1)/6 = k(k + 2)(2k + 3)/6.
(e) By the inductive hypothesis, we know
12 + 22 + · · · + k 2 + (k + 1)2 = k(k + 1)(2k + 1)/6 + (k + 1)2 = (∗),
then writing the right hand side as a single fraction, we find
(∗) = (k(k + 1)(2k + 1) + 6(k + 1)2 )/6 = (k + 1)(k(2k + 1) + 6(k + 1))/6
= (k + 1)(2k 2 + 7k + 6)/6 = (k + 1)(k + 2)(2k + 3)/6
(f) This proves the statement, since part (b) shows the statement is true for n = 1, thus
part (d) shows the statement is true for n = 2, therefore it is true for n = 3, therefore it
is true for n = 4, etc.
(15)
• Base case: If n = 1, then the left hand side is 1 · 2 = 2, and the right hand side
is 1(1 + 1)(1 + 2)/3 = 2. Therefore the statement is true for n = 1.
• Inductive step: Suppose for some k ≥ 1,
1 · 2 + 2 · 3 + · · · + k(k + 1) = k(k + 1)(k + 2)/3.
(1)
Then, we can write 1 · 2 + 2 · 3 + · · · + k(k + 1) + (k + 1)(k + 2) = k(k + 1)(k + 2)/3 +
(k + 1)(k + 2) = (∗). Then, combining the right hand side into a single fraction, we find
(∗) = (k(k + 1)(k + 2) + 3(k + 1)(k + 2))/3 = (k + 1)(k + 2)(k + 3)/3,
(2)
which proves the statement.
(21)
• Base case: The smallest integer strictly greater than 4 is 5, so let n = 5. Then, 25 = 32
and 52 = 25. Thus, 25 > 52 .
• Inductive step: Suppose for some k > 4, we have 2k > k 2 . Then, (k+1)2 = k 2 +2k+1 <
k 2 + k 2 = 2k 2 and by the inductive hypothesis, 2k 2 < 2 · 2k = 2k+1 .
(28)
• Base case: If n = 3, then 32 − 7 · 3 + 12 = 0.
• Inductive step: Suppose for some k ≥ 3, k 2 − 7k + 12 ≥ 0. Then,
(k + 1)2 − 7(k + 1) + 12 = k 2 − 7k + 12 + 2k − 6 ≥ 2k − 6,
by the inductive hypothesis. As an increasing linear function, 2k − 6 ≥ 0 for all k ≥ 3.
Therefore, (k + 1)2 − 7(k + 1) + 12 ≥ 0.
1
Section 4.2
(7) Claim: We can form the dollar amounts $0, $2, and every amount ≥ $4 using only $2 and
$5. Clearly $0 and $2 satisfy the claim.
• Base case: Observe 4 = 2 · 2 and 5 = 5 · 1 both satisfy the claim.
• Inductive step: Suppose for all 5 ≥ k < n, k can be written as k = 2a + 5b for some
non-negative integers a, b. Then, 4 ≥ n−2 so by the base case or the induction hypothesis
we can write n − 2 = 2a + 5b for some non-negative a, b. Then, n = 2(a + 1) + 5b.
(12)
• Base case: If n = 1, then 1 = 20 is written as a power of 2.
• Inductive step: Suppose for all k < n, k can be written as a sum of distinct powers
of 2. If n is even, then n = 2k for some k < n, therefore by the inductive hypothesis,
we can write k = 2a1 + 2a2 + · · · + 2am for some ai 6= aj for all i 6= j. Then, n =
2a1 +1 +2a2 +1 +· · ·+2am +1 . On the other hand, if n is odd, then n = k +1, and k satisfies
the conditions of the inductive hypothesis so again we can write k = 2a1 +2a2 +· · ·+2am .
Then, n = 2a1 + 2a2 + · · · + 2am + 1 = 2a1 + 2a2 + · · · + 2am + 20 . Since n is odd, k is
even, so ai 6= 0 for any i.
Section 4.3
(3) See solution in book.
(13) Recall the Fibonacci number fn = fn−1 + fn−2 , beginning with f0 = 0 and f1 = 1.
• Base case: If n = 1, then f1 = 1 and f2 = f1 + f0 = 1 + 0 = 1.
• Inductive step: Suppose for some k ≥ 1, f2k = f1 + f3 + · · · + f2k−1 . Then, f2(k+1) =
f2(k+1)−1 + f2(k+1)−2 = f2k+1 + f2k . By the inductive hypothesis,
f2(k+1) = f2k+1 + f2k = f2k+1 + f2k−1 + · · · + f3 + f1 .
2