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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
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 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
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