Question

Prove that the sum of the first n odd positive integers is n2.

In other words, show that 1 + 3 + 5 + .... + (2n + 1) = (n + 1)2
for all n ∈ N.

Answer 1

To prove that the sum of the first n odd positive integers is n squared, we rearrange the terms such that each pair sums to a multiple of n and then apply the sum formula.

Step-by-step explanation:

To prove that the sum of the first n odd positive integers is n2, we use a simple rearrangement method. The sequence of odd numbers up to the nth term is 1, 3, 5, ..., (2n - 3), (2n - 1), and their sum can be written as S = 1 + 3 + 5 + ... + (2n - 3) + (2n - 1). Now, rewrite this by taking (n - 1) from the last term and adding it to the first term, which gives S = 2[1 + (n - 1)] + 3 + ... + (2n - 3) + [n + (2n - 1) - (n - 1)]. This simplifies as S = 2[n] + 3 + ... + (2n - 3) + n. Then take (n - 3) from the second to last term and add to the second term, and continue this process until all terms are 'n'. This results in S = 2[n + n + ... + n + n], where there are n terms of n, thus S = 2n2. To get the correct sum for the sequence, divide by 2, so S = n2.