April 8th, 2014, 03:01 AM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  induction
Use the method of induction to prove that the sum of first n odd numbers is equal to n^2 My attempt: for n = 1 1 = 1^2 = 1 So it is true for n = 1 Assume true for 1 + 3 + 5 +...+ 2n  1 = n^2 Now try for (n + 1)th term 1 + 3 + 5 +...+ 2n  1 + 2(n + 1) 1 = n^2 + 2(n + 1)  1 = n^2 + 2n + 2  1 = n^2 + 2n + 1 = (n + 1)^2 So it is true for 1, n and n + 1 Therefore the sum of first n odd numbers = n^2 Is this the correct method? Am I missing something? thanks 
April 8th, 2014, 03:26 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra 
You have not missed anything, your proof is perfect. 
April 8th, 2014, 03:27 AM  #3 
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 
The proof is correct, but there is no reason to represent 2n + 1 as 2(n + 1)  1. Simply 1 + 3 + 5 +...+ (2n  1) + (2n + 1) = n^2 + 2n + 1 (by IH) = (n + 1)^2. Also, it's a good idea to indicate where exactly the induction hypothesis is used. 

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