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 163
Performing several simplifications yields:
l^ + 2^ + 3^ + --- + n^ + in+lf
= (w + l)^fe + (n+l)
2 (n^ + 4n + 4)
(n+lf[in+l)
+ lf
Therefore, by the principle of mathematical induction, the given
formula holds true for all positive integer numbers.
22. (The two numbers a and b appear in a formula. Therefore, we can try
to manipulate the formula to obtain exphcit information about them.)
Because ab = {a^ + lab + b^)/4, we have 4ab = a^ -\- lab + b^. Thus,
0 = a^ - lab + b^. The right-hand side of the equahty is equal to
(a — b)^. Therefore, we obtain (a — bf = 0. This implies that
a — b = 0. Thus, we can conclude that a = b.
23. (We can start working on the given equation, which involves the two
numbers a and b, in the hope of obtaining useful clues about them.)
From the equahty:
we obtain:
lab = a^ + lab + b^.
Therefore,
0 = a' + b\
Because a^ and b^ are both nonnegative numbers, their sum can be
equal to zero if and only if they are both equal to zero (because
cancellation is not possible). However, a^ = 0 and b^ = 0 implies
a = b = 0. Therefore, the statement is true.
24. We are going to prove the given statement using mathematical
induction.