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Unformatted text preview: H om e w o r k 6
Sol u t ion
P r o b l e m 1 Se c t io n 4.1, E x e r c ise 4, p age 279
a) b) Both sides of P(1) shown in part (a) equal 1.
c) d) For each, k 1 that P(k) implies P(k+1) e) ! f) We have completed both the basis step and inductive step so by the principle of mathematical
induction the statement is true for every positive integer n. P r o b l e m 2 Se c t io n 4.1, E x e r c ise 6, p age 280
Proof) Let P(n) be 1)Basic step: P(1) is true because 1 1!=1=(1+1)!1
2) Inductive step: Assume that P(k) is true. Then, (1 1!+2 2!+ +k k!)+(k+1)(k+1)!=(k+1)!1+(k+1)(k+1)!=(k+1)!(1+k+1)1=(k+2)(k+1)!1
=(k+2)!1=P(k+1)
We have completed both the basis step and inductive step so by the principle of mathematical
induction the statement is true for every positive integer n. P r o b l e m 3 Se c t io n 4.1, E x e r c ise 14, p age 280
Proof) Let P(n) be 1) Basic step: P(1)=1 21=(11)21+1+2=2
2) Inductive step: Assume P(k) is true. Then,
(1 21+2 22+ +k 2k)+(k+1) 2k+1=(k1) 2k+1+2+(k+1) 2k+1
=2k+1 (k1+k+1)+2=2k 2k+1 +2=k 2k+2 +2=P(k+1)
We have completed both the basis step and inductive step so by the principle of mathematical
induction the statement is true for every positive integer n. P r o b l e m 4 Se c t io n 4.2, E x e r c ise 4, p age 291
a) P(18) is true, because we can form 18 cents of postage with one 4cent stamp and two 7cent
stamps. P(19) is true, because we can form 19 cents of postage with three 4cent stamps and one
7cent stamp. P(20) is true, because we can form 20 cents of postage with five 4cent stamps.
P(21) is true, because we can form 21 cents of postage with three 7cent stamps.
b) Inductive hypothesis: the statement that using 4cent and 7cent stamps we can form j cents
postages for all j with 18 j k, where we assume k 21.
c) Assuming the inductive hypothesis, we can form k+1 cents postage using just 4cent and 7cent stamps.
d) Because k 21, we know that P(k3) is true, that is, that we can form k3 cents of postage.
Put one more 4cent stamp on the envelope, and we have formed k+1 cents of postage. e) We have completed both basis step and the inductive step, so by the principle of strong
induction, the statement is true for every integer n greater than or equal to 18. P r o b l e m 5 Se c t io n 4.2, E x e r c ise 12, p age 292
Proof)
1)Basic step:
1 = 20, 2 = 21, 3 = 21 + 20, 4 = 22, 5 = 22 + 20, and so on.
2) Inductive step: Assume the inductive hypothesis that every positive integer up to k can be
written as a sum of distinct powers of 2. We must show that k + 1 can be written as a sum of
distinct powers of 2. If k + 1 is odd, then k is even, so 20 added. If k + 1 is even, then (k + 1)/2
is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct
powers of 2. Increasing each exponent by 1 doubles the value and gives us the desired sum for
k + 1. P r o b l e m 6 Se c t io n 4.2, E x e r c ise 30, p age 293
In the inductive step, when k=0, a0=1 for all nonzero a, but a1 is not one. In the denominator, we have , but 1 is negative, so it's not covered under our inductive assumption. Therefore, the inductive step fails. ...
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This note was uploaded on 11/05/2011 for the course CSCE 222 taught by Professor Math during the Spring '11 term at Texas A&M.
 Spring '11
 math

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