Science:Math Exam Resources/Courses/MATH220/December 2011/Question 06 (a)/Solution 1

We start our induction at n = 1. In that case we have that 2n - 1 = 1 and so the sum is just 1 alone which is a perfect square.

We can check n = 2 if we want as well, in that case 2n - 1 = 4 - 1 = 3 and so the sum is 1 + 3 = 4 which is the square of 2.

For n = 3 we have that 1 + 3 + 5 = 9 which is the square of 3; we now see something which seems to be a pattern, that is for n = k the sum will end up being the square of the number k. And so we conjecture that

$\displaystyle 1+3+5+\ldots +(2n-1)=n^{2}$

We have proved above that this is true for n = 1, 2 and 3. We will use induction to prove that is holds for all values of n. Since we have the first step(s) done, we now assume that the conjecture holds for n = k and we will show that it still holds for n = k+1. Indeed

${\begin{aligned}\underbrace {1+3+5+\ldots +(2k-1)} _{=k^{2}}+(2(k+1)-1)&=k^{2}+(2k+2-1)\\&=k^{2}+2k+1\\&=(k+1)^{2}\end{aligned}}$

This shows that our conjecture is true for n = k+1 and thus concludes our proof.