Beatley324
3

# Find the sum of a finite geometric sequence from n = 1 to n = 7, using the expression −4(6)n − 1.

$\bf \textit{sum of a finite geometric sequence} \\\\ S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term in the sequence}\\ r=\textit{common ratio} \end{cases}\\\\ -----------------------------\\\\ \textit{the }n^{th}\textit{ term of a geometric sequence} \\\\ a_n=a_1\cdot r^{n-1}\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term in the sequence}\\ r=\textit{common ratio} \end{cases}\\\\ -----------------------------$ $\bf \textit{now, let's see yours} \\\\ -4(6)^{n-1}\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term in the sequence}\to &-4\\ r=\textit{common ratio}\to &6 \end{cases}$ so.. now we know what the common ratio is, 6, and what the first term is, -4, and our nth term is 7, since we're asked to do the sum from 1 to 7 so... let us use that in the SUM for a finite sequence $\bf S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\implies S_7=-4\left( \cfrac{1-(6)^7}{1-(6)} \right)$