## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\sum_{k=1}^{12} (2k-1)$
We need to compute summation formula for the sequence $1+3+5+7+.......+[2(12)-1]$ When $k=1$, the first term is $1=(2)(1)-1$ and for $k=2$ , the second term is $3=(2)(2)-1$. This information indicates the $k^{th}$ term as $(2k-1)$. So, we write the the summation formula for $k^{th}$ term as follows: $1+3+5+7+.......+[2(12)-1]= \sum_{k=1}^{12} (2k-1)$