Question

10. Find the 9th partial sum of the summation of 3i minus 4, from i equals 1 to infinity.

102
99
205
200

Answer 1

@mathmale @jigglypuff314

Answer 2

For the sum\[\sum_{i=1}^{n}3i-4\]
the partial sum is as follows
\[\sum_{i=1}^{n}3i-4=\frac{ 3 }{ 2 }n^2-\frac{ 5 }{ 2 }n\]
So the 9th partial sum is just n=9
\[\frac{ 3 }{ 2 }9^2-\frac{ 5 }{ 2 }9=99\]