Write the sum $LaTeX: \displaystyle 41+45+49 \ldots +189+193$ in sigma notation and then find the sum.

The common difference is given by $LaTeX: \displaystyle a_2-a_1=45-(41)=4$ . Using the first term gives the sequene $LaTeX: \displaystyle a_n= 41+(n-1)(4)$ . Setting the general term equal to the last term and solving for $LaTeX: \displaystyle n$ gives $LaTeX: \displaystyle 41+(n-1)(4)=193 \implies n = 39$ . Writing in sigma notation gives $LaTeX: \displaystyle \displaystyle \sum_{n=1}^{39} \left(4 n + 37\right)$ . Using the formula for a finite arithmetic sum gives $LaTeX: \displaystyle \frac{ 39(41+193) }{2}=4563$ .