The given sum is
[tex]\sum ^9_{k\mathop=4}(5k+3)[/tex]
This means we have to replace k = 4, 5, 6, 7, 8, 9, and then we sum
[tex]\begin{gathered} (5\cdot4+3)+(5\cdot5+3)+(5\cdot6+3)+(5\cdot7+3)+(5\cdot8+3)+(5\cdot9+3) \\ 20+3+25+3+30+3+35+3+40+3+45+3=213 \\ \end{gathered}[/tex]
Hence, the sum is equal to 213. The right answer is C.
