## College Algebra (10th Edition)

$1+3+5+7+\displaystyle \cdots+[2(12)-1]=\sum_{k=1}^{12}(2k-1)$
$1+3+5+7+\cdots+[2(12)-1]=$ There are 12 terms. Count them with index k: $\displaystyle \sum_{k=1}^{13}$(...) The first term (when k=1) is $1=2(1)-1$, The second term (when k=2) is $3=2(2)-1$, The kth is $=2(k)-1$ so $1+3+5+7+\displaystyle \cdots+[2(12)-1]=\sum_{k=1}^{12}(2k-1)$