Examples 2.3.4(a):
Use induction to prove the following statements:
1. The sum of the first n positive integers is n (n+1) / 2
 The sum of the first n positive integers is n (n+1) / 2.
 If a, b > 0, then (a + b)^{n} a^{n} + b^{n} for any positive integer n.
To use induction, we have to proceed in fours steps.
 Property Q(n):
 1 + 2 + 3 + ... + n = n (n+1) / 2
 Check Q(1):
 1 = 1 * 2 / 2, which is true.
 Assume Q(n) is true:
 Assume that 1 + 2 + 3 + ... + n = n (n+1) / 2.
 Check Q(n+1):

1 + 2 + 3 + ... + n + n + 1 =
Hence, Q(n+1) is true under the assumption that Q(n) is true.
= (1 + 2 + 3 + ... + n) + (n + 1) =
= n (n+1) / 2 + (n+1) =
= n (n+1) / 2 + (2n + 2) / 2 =
= (n+1)(n+2) / 2
2. If a, b > 0, then (a + b)^{n} a^{n} + b^{n} for any positive integer n.
Using an induction proof, we have to proceed in four steps:
 Property Q(n):
 (a + b)^{n}
a^{n} + b^{n}
if a, b > 0
 Check Q(1):
 (a + b) a + b, which is true.
 Assume Q(n) is true:
 Assume that
(a + b)^{n}
a^{n} + b^{n}
 Check Q(n+1):

(a + b)^{n + 1} = (a + b)^{n} (a + b) (a^{n} + b^{n}) (a + b) =
because a, b > 0. Hence, Q(n+1) is true under the assumption that Q(n) is true.
= a^{n + 1} + a^{n} b + a b^{n} + b^{n + 1} a^{n + 1} + b^{n + 1}