# Prove the identity 1+3+5+...+(2n-1)=n^2

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### 2 Answers

For demonstrating the identity, we'll use the method of mathematical induction, which consists in 3 steps:

1) verify that the method works for the number 1;

2) assume that the method works for an arbitrary number, k;

3) prove that if the method works for an arbitrary number k, then it work for the number k+1, too.

4) after the 3 steps were completed, then the formula works for any number.

Now, we'll start the first step:

1) 1=1^2 => 1=1 true.

2) 1+3+5+...+(2k-1)=k^2 , true.

3) If 1+3+5+...+(2k-1)=k^2, then

1+3+5+...+(2k-1)+(2k+2-1)=(k+1)^2

Let's see if it is true.

For the beginning, we notice that the sum from the left contains the assumed true equality, 1+3+5+...+(2k-1)=k^2. So, we'll re-write the sum by substituting 1+3+5+...+(2k-1) with k^2.

k^2 + (2k+1) = (k+1)^2

We'll open the brackets:

k^2 + 2k+1 = k^2 + 2k+1 true.

4) The 3 steps were completed, so the identity is tru for any value of n.

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

1+3+5+...2n-1 is an A.P wirh starting rem a =1 and common ratio 2. and the number of terms, n.

So the sum is Sn = [a+a+(n-1)d}n/2

=[1+1+(n-1)*2]n/2

=(1+1+2n-2)n/2

2n^2/2

=n^2