Prove for every natural number n, 1+3+5+...(2n-1)=(n^2)

Prove for every natural number n, 1+3+5+...(2n-1)=(n^2)

First let us consider an example where n=6, then we see that

1+3+5+7+9+11 = 36 = (6^2) = (n^2)

Now we have use to the principle of mathematical induction (PMI) to prove that this is true for all cases.

So we define a set S to be the set of natural numbers for which the statment:

for every natural number n, 1+3+5+...(2n-1)=(n^2)

is true.

First, let us find that the statement is true for the first natural number: 1=(1^2)

Thus, 1 is an element in our set S.

Now that we have proved the case for 1, we should prove the same thing for the next case which is (n+1), if we can prove that (n+1) is in the set S, then by induction we have proved it to be true for all n.

To do this we take the original antecedent:

1+3+5+...(2n-1)=(n^2) and add 2(n+1) - 1 to both sides, doing that we get

1+3+5+...(2n-1) + [2(n+1) - 1] = (n^2) + 2(n+1) - 1

= (n^2) + 2n + 1
= (n+1)^2

thus (n+1) is in our set S, and the statement: 1+3+5+...(2n-1)=(n^2), is true for all n.

Another way to think about a proof by induction is as a direct proof in which we are saying, for all n, n in S implies that n+1 is in S, and thus is true for all n.

~~~~