Prove by induction that the sum of the first n positive odd integers is n^(2).

Question

Prove by induction that the sum of the first  n positive odd integers is  n^(2).

Bytelearn · Accepted Answer

Base Case of Induction: Let's start with the base case of induction, which is n = 1. The first positive odd integer is 1. We need to check if the sum of the first odd integer equals 1^2. Sum for n = 1: 1 1^2 = 1 Since both are equal, the base case holds true.Assumption for Positive Integer: Now, let's assume that the statement is true for some positive integer k, which is our induction hypothesis. This means we assume the sum of the first k positive odd integers is k^2. Sum for n = k: 1 + 3 + 5 + ... + (2k - 1) = k^2 This is our assumption and we will use it to prove the statement for n = k + 1.Proving Statement for n = k + 1: Next, we need to show that if the statement is true for n = k, then it must also be true for n = k + 1. The sum of the first k + 1 positive odd integers would be the sum of the first k odd integers plus the next odd integer, which is 2(k + 1) - 1. Sum for n = k + 1: 1 + 3 + 5 + ... + (2k - 1) + [2(k + 1) - 1] Using our induction hypothesis, we can replace the sum of the first k odd integers with k^2. So, Sum for n = k + 1: k^2 + [2(k + 1) - 1]Simplifying the Expression: Now, let's simplify the expression for the sum of the first k + 1 odd integers. Sum for n = k + 1: k^2 + [2k + 2 - 1] Sum for n = k + 1: k^2 + 2k + 1 Sum for n = k + 1: (k + 1)^2 This shows that the sum of the first k + 1 positive odd integers is indeed (k + 1)^2, which is what we needed to prove.

Prove by induction that the sum of the first $n$ positive odd integers is $n^{2}$ .

Full solution

More problems from Positive and negative square roots

Related topics

Prove by induction that the sum of the first nnn positive odd integers is n2n^{2}n2.

Prove by induction that the sum of the first $n$ positive odd integers is $n^{2}$ .