Sum of odd numbers/Induction/Exercise/Solution

An odd number has the form ${}2k-1$, the sum of the first ${}n$ odd numbers equals hence

$\sum _{k=1}^{n}(2k-1).$

We claim that this equals ${}n^{2}$. For ${}n=1$ the statement is true, since the sum is just ${}1=1^{2}$. Suppose now that the statement is known for some ${}n$. Then

${}\sum _{k=1}^{n+1}(2k-1)=\sum _{k=1}^{n}(2k-1)+2n+1=n^{2}+2n+1=(n+1)^{2}\,.$

To the exercise