Prove by induction that the sum from r=1 to n of (2r-1) is equal to n^2.

Firstly, we note that there are 4 key steps for any proof by induction. (1) Basis : check the statement works for n=1. (2) Assumption : assume the statement is true for n=k. (3) Induction : prove the statement is true for n=k+1 using our assumption. (4) Conclusion : conclude what we have shown. For step 1 of this question, we want to make sure that the left hand side is equal to the right hand side. If we let n=1, then for the left hand side we have the sum from r=1 to 1 of (2r-1) which is just one term where r=1 so we have 2(1)-1=1. Now we need to check that the statement is true: on the right hand side we have n2, so when n=1 we have 12=1. Hence, we know that the statement is true for n=1 because the left hand side and right hand side of the equation statement are the same. Now for step 2 we assume that the sum from r=1 to k of (2r-1) is equal to k2. For step 3, we need to think about when n=k+1.If we want the sum from r=1 to k+1 of (2r-1), we can think of this as being the sum from r=1 to k, plus the final term which is (2(k+1)-1). Using our assumption and expanding the final term, we can rewrite this as k2+2k+1, then factorise it to get (k+1)2 which is what we wanted to prove. To conclude the proof, we need to say that if the statement is true for n=k, then we have shown it is also true when n=k+1. Because the statement was true for n=1, we know that it is true for any n value that is a positive integer.

JL

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