Using mathematical induction, prove that the sum of the first 𝒏 odd positive integersis equal to 𝒏𝟐
Question
Using mathematical induction, prove that the sum of the first n odd positive integers is equal to n².
Solution
Sure, here is the proof using mathematical induction:
Step 1: Base Case Let's start with n=1. The first odd positive integer is 1, and 1^2 = 1. So, the statement holds true for n=1.
Step 2: Inductive Step Assume the statement is true for some positive integer k. That is, the sum of the first k odd positive integers is equal to k^2.
Step 3: Inductive Hypothesis We need to prove that the statement is true for n=k+1. That is, the sum of the first k+1 odd positive integers is equal to (k+1)^2.
The sum of the first k+1 odd positive integers can be written as (sum of the first k odd positive integers) + (k+1)th odd positive integer.
By the inductive hypothesis, the sum of the first k odd positive integers is k^2. The (k+1)th odd positive integer is 2k+1.
So, the sum of the first k+1 odd positive integers is k^2 + 2k + 1.
Simplifying, we get (k+1)^2.
Therefore, by mathematical induction, the sum of the first n odd positive integers is equal to n^2 for all positive integers n.
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