16
\(\sum (-11+4i)\)
i=8
∑(-11+4i), from i=8 to 16
Sum(i=8)^16(4 i-11) = 333
= (4(8)-11)+(4(9)-11)+(4(10)-11)+(4(11)-11)+(4(12)-11)+(4(13)-11)+(4(14)-11)+(4(15)-11)+(4(16)-11)
= 280
Oh is that how you do it when i dosnt equal one i thought you had to go by 8ths. that explains a lot thank you so much
+9589 
+116 
