Question

Find the sum of a finite geometric sequence from n=1 to n=8, using the expression -2(3)^n-1

Answer 1

$\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio} \end{cases} \\\\\\ \sum\limits_{n=1}^{8}~-2(3)^{n-1}~~ \begin{cases} n=8\\ a_1=-2\\ r=3 \end{cases}\implies S_8=-2\left( \cfrac{1-3^8}{1-3} \right) \\\\\\ S_8=\underline{-2}\left( \cfrac{1-6561}{\underline{-2}} \right)\implies s_8=1-6561\implies S_8=-6560$