Answer
$1+3+5+7+\cdots+(2n-1)$
Work Step by Step
There are n terms, as the index k changes from 0 to n-1.
The index k dictates how the terms are formed:
$\displaystyle \sum_{k=0}^{n-1}(2k+1)=\\=(2(0)+1)+(2(1)+1)+(2(3)+1)+...+(2(n-1)+1)$
$=1+3+5+7+\cdots+(2n-1)$