Question

what is the sum of first n odd positive integers​

Answer 1

Answer:

Step-by-step explanation:

Principle of Mathematical Induction. Let P be a property of positive integers such that:

1. Basis Step: P(1) is true, and

2. Inductive Step: if P(n) is true, then P(n + 1) is true.

Then P(n) is true for all positive integers.

Remark: The premise P(n) in the inductive step is called Induction

Hypothesis.

The validity of the Principle of Mathematical Induction is obvious.

The basis step states that P(1) is true. Then the inductive step implies

that P(2) is also true. By the inductive step again we see that P(3)

is true, and so on. Consequently the property is true for all positive

integers.

Remark: In the basis step we may replace 1 with some other integer

m. Then the conclusion is that the property is true for every integer n

greater than or equal to m.

Example: Prove that the sum of the n first odd positive integers is

n2, i.e., 1 + 3 + 5 + · · · + (2n − 1) = n2.

Answer : Let S(n) = 1 + 3 + 5 + · · · + (2n − 1). We want to prove

by induction that for every positive integer n, S(n) = n2.

1. Basis Step: If n = 1 we have S(1) = 1 = 12, so the property is

true for 1.

2. Inductive Step: Assume (Induction Hypothesis) that the property is true for some positive integer n, i.e.: S(n) = n2. We must

prove that it is also true for n + 1, i.e., S(n + 1) = (n + 1)2. In

fact:

S(n + 1) = 1 + 3 + 5 + · · · + (2n + 1) = S(n) + 2n + 1 .

4.2. MATHEMATICAL INDUCTION 63

But by induction hypothesis, S(n) = n2, hence:

S(n + 1) = n2 + 2n + 1 = (n + 1)2 .

This completes the induction, and shows that the property is true for

all positive integers.

Example: Prove that 2n + 1 ≤ 2n for n ≥ 3