look for S9: upper limit= 9, nth term(explicit)=(2k-3), k(index)=1(lower limit)

Is this what you mean? I would assume so.

\sum_{k=1}^{9}(2k-3)

Just add up the terms. It's not a difficult problem.

(2(1)-3)+(2(2)-3)+(2(3)-3)+... +(2(9)-3)=?