Mathematics - Sequence & Series Question with Solution | TestHub

Row $i$ has $t_i=2i-1$ terms, so ${\sum }_{i=1}^n{t}_i=1+3+5+\cdots +(2n-1)=n^2$. The last term of row $n-1$ is $(n-1{)}^2$, so the first term of row $n$ is $(n-1{)}^2+1$. For $n=15$ this is $196+1=197$. ZZZ DUMMY - DO NOT IMPORT \end{document}