epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Solutions for the Exercises at the End of the Sections and the Review Exercises 135
When we use the inductive hypothesis, the equaUty becomes:
9^+1 „ I ^ 9(9" _ 1) 4- 8
= 9(8^)4-8 = 8(9^ + 1).
Because 9g + 1 is an integer number, it follows that 9"+^ ~ 1 is divisible
by 8. Thus, by the principle of mathematical induction, the statement
is true for all fc > 1.
There is another way to construct a proof without using mathematical
induction. The basic tool is the factorization formula for the difference
of two powers:
9^^ - 1 = 9^ - 1^
= (9 - l)(9'^-i + 9^-^ + • • • + 1)
= 8(9^^-^ + 9^-2 + ... 4-1) = 85
where s is an integer number. The formula used above requires that
/c> 1. So, we need to use a separate proof for /c= 1. When /c= 1, we
have 9^ — 1 = 8, which is divisible by 8.
3. (i) Let us check whether the statement is true for k=l. When /c= 1,
2k = 2. Therefore, we have only one number in the left-hand side of
the equation. We obtain 2= 1^ + 1, which is a true statement,
(ii) Assume that the statement is true for k = n; that is, 2 + 4 +
6 H \-2n = n^ -\-n.
(iii) Prove that the equality is true for k = n-\-l. The last number in
the left-hand side is 2{n +1) = 2n + 2. So, we need to add all the
even numbers between 2 and 2n + 2. The largest even number
smaller than 2n-\-2 is {2n-\-2) — 2 = 2n, because the difference
between two consecutive even numbers is 2. Thus, we need to
prove that 2 + 4 + 6 + • • • + 2n H- (2n + 2) = (n + 1)^ + (n -f-1).
Using the associative property of addition and the inductive
hypothesis, we obtain:
2 + 4 + 6 + . • • + 2n + (2n + 2)
= [2 + 4 + 6 + ..• + 2n] + (2n + 2)
= [n^ + n] + (2n + 2)
= (n^ + 2n+l) + (n+l)
= (n+l)2 + (n+l).
Thus, by the principle of mathematical induction, the given equahty
holds true for all /c> 1.