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 137
Using the associative property of addition and the inductive
hypothesis, we obtain
n+l
^•••n^)+G ^•+fi + •2
©
l\ «+i
l_(l/2)«+i ^ /i\"+i
l-(l/2)
l-(l/2)"+^+(l/2)"+^-(l/2)"
l-(l/2)
1 - (1/2)"+^
-1.
l-(l/2)
So, by the principle of mathematical induction, the statement is true
for all k>\.
6. a. Check the inequality for k — 'i.ln this case 3^ = 9 and 5(3!) = 5(6) =
30. So the statement is true.
b. Let us assume that the inequality holds true for an arbitrary k = n;
that is, assume that 5n! > n^.
c. Is (n +1)^ < 5(n +1)!? By the properties of factorials,
(n+l)! = (n+l)n!. So
5(n + 1)! = 5(n + l)n! = 5nn! + 5n!
> 5nn! + n^.
The fact that n > 3 impUes that n! > 3. So 5nn! > 15n. Thus,
5(« + 1)! > 5nn! + n^
> n^ + 15n> n^ + 3n
= n^ + 2n + n > n^ + 2n + 1
= (n+l)l
So, by the principle of mathematical induction, the inequality is true
for all M > 3.
7. a. The statement is true for n = 1 because 1^ — 1 = 0, and 0 is divisible
by 4.
b. Assume the statement is true for all the odd numbers from 1 to n,
where n is odd. In particular, assume that n^ — 1 = 4m for some
positive integer m.
c. Is the statement true for the next odd number (namely, n-\-2)l
Observe that (n 4- if - 1 ^(n^ - 1) + 4(n + 1). Because n^ - 1 =4m
for some positive integer m, by the inductive hypothesis we have
{n-\-2f' — I = 4(m + n + 1). Because the number m + n + 1 is an
-1