Question

Find the sum of the 3rd through 14th terms of the series.

A(n)=4(3)^n−1

Answer 1

`\qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle 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=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio} \end{cases}`

let's get the summation of the first 14 terms and then the summation of the first 2 terms, and subtract the later from the former, leaving us only with the summation from 3 through 14.

`A(n)=\stackrel{a_1}{4}(\underset{r}{3})^(n-1)\implies \displaystyle\sum_(i=1)^(14)~~4(3)^(n-1)~~ -~~\sum_(i=1)^(2)~~4(3)^(n-1) \\\\\\ 4\left( \cfrac{1-3^(14)}{1-3} \right)~~ - ~~4\left( \cfrac{1-3^(2)}{1-3} \right)\implies 4(2391484)~~ - ~~4(4) \\\\\\ 9565936~~ -~~16\implies \boxed{9565920}`