Question

Find the 7th partial sum of summation of 6 open parentheses 3 close parentheses to the I minus 1 power from 1 to infinity

Answer 1

Is this your series?
\[S=\sum_{\ell=1}^\infty 6\times3^{\ell-1}\]
If so, the seventh partial sum is
\[S_7=\sum_{\ell=1}^7 6\times3^{\ell-1}=6\sum_{\ell=1}^7 3^{\ell-1}\]which can be computed with the geometric sum formula,
\[\sum_{k=1}^n r^{k-1}=\begin{cases}\dfrac{1-r^{n}}{1-r}&\text{for }r\neq1\\[1ex]n&\text{for }r=1\end{cases}\]Clearly \(S_7\) uses \(r=3\) and \(n=7\).