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

Special Kinds of Theorems 57
If we simplify the right-hand side of the inequahty, we can rewrite it as
(n + 1)^ > 3n^ + 9^ + 7.
By finding the third power of the binomial n +1 and using the
inductive hypothesis to replace n^ we obtain
(n-\-lf
= n^ + 3n^ + 3n+l
> (3n^ + 3n + 1) + 3n^ + 3n + 1.
Combining some of the similar terms yields
(n + 1)^ > 3n^ + 3n^ + 6n + 2.
At this point, there are several ways to proceed. Thus, it is very
important to keep in mind the conclusion we want to reach. Because
n>4,3n^ = 3nxn> I2n. Therefore,
{n + if > 3n^ + 3n^ + 6w + 2
Again, n > 4 implies 9n + 2 > 7. So,
> 3n^ + 12n + 6n + 2
= 3n^ + 9n + (9n + 2).
(n + 1)^ > 3n^ + 9n + 7.
This is exactly the conclusion we wanted to reach. Therefore, by the principle
of mathematical induction the inequahty holds true for all integers greater
than or equal to 4. •
EXERCISES
Prove the following statements:
1. For all positive integers /c,
1 + 2 + 2^ + 2^ +••• + 2^-^ =2^-1.
2. The number 9^ - 1 is divisible for all fe > 1.
3. For all integer numbers fc > 1
2 + 4 + 6 + ... + 2/c = /c^ + /c.