\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Write the sum in sigma notation \(\displaystyle 15-\frac{45}{2}+\frac{135}{4}\ldots\) and then find the sum.
Taking the ratio of the first two terms gives \(\displaystyle r = \frac{a_2}{a_1}=\frac{- \frac{45}{2}}{15}=- \frac{3}{2}\). Using \(\displaystyle a_{1}\) and \(\displaystyle r\) to write in sigma notation gives \(\displaystyle \sum_{n=1}^{\infty}15\left( - \frac{3}{2} \right)^{n-1}\). Since \(\displaystyle |r| = \frac{3}{2} \geq 1\) the geometric series diverges.
\begin{question}Write the sum in sigma notation $15-\frac{45}{2}+\frac{135}{4}\ldots$ and then find the sum.
\soln{9cm}{Taking the ratio of the first two terms gives $r = \frac{a_2}{a_1}=\frac{- \frac{45}{2}}{15}=- \frac{3}{2}$. Using $a_{1}$ and $r$ to write in sigma notation gives $\sum_{n=1}^{\infty}15\left( - \frac{3}{2} \right)^{n-1}$. Since $|r| = \frac{3}{2} \geq 1$ the geometric series diverges.}
\end{question}
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}
\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}
\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}
\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Write the sum in sigma notation <img class="equation_image" title=" \displaystyle 15-\frac{45}{2}+\frac{135}{4}\ldots " src="/equation_images/%20%5Cdisplaystyle%2015-%5Cfrac%7B45%7D%7B2%7D%2B%5Cfrac%7B135%7D%7B4%7D%5Cldots%20" alt="LaTeX: \displaystyle 15-\frac{45}{2}+\frac{135}{4}\ldots " data-equation-content=" \displaystyle 15-\frac{45}{2}+\frac{135}{4}\ldots " /> and then find the sum.</p> </p><p> <p>Taking the ratio of the first two terms gives <img class="equation_image" title=" \displaystyle r = \frac{a_2}{a_1}=\frac{- \frac{45}{2}}{15}=- \frac{3}{2} " src="/equation_images/%20%5Cdisplaystyle%20r%20%3D%20%5Cfrac%7Ba_2%7D%7Ba_1%7D%3D%5Cfrac%7B-%20%5Cfrac%7B45%7D%7B2%7D%7D%7B15%7D%3D-%20%5Cfrac%7B3%7D%7B2%7D%20" alt="LaTeX: \displaystyle r = \frac{a_2}{a_1}=\frac{- \frac{45}{2}}{15}=- \frac{3}{2} " data-equation-content=" \displaystyle r = \frac{a_2}{a_1}=\frac{- \frac{45}{2}}{15}=- \frac{3}{2} " /> . Using <img class="equation_image" title=" \displaystyle a_{1} " src="/equation_images/%20%5Cdisplaystyle%20a_%7B1%7D%20" alt="LaTeX: \displaystyle a_{1} " data-equation-content=" \displaystyle a_{1} " /> and <img class="equation_image" title=" \displaystyle r " src="/equation_images/%20%5Cdisplaystyle%20r%20" alt="LaTeX: \displaystyle r " data-equation-content=" \displaystyle r " /> to write in sigma notation gives <img class="equation_image" title=" \displaystyle \sum_{n=1}^{\infty}15\left( - \frac{3}{2} \right)^{n-1} " src="/equation_images/%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D15%5Cleft%28%20-%20%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%5E%7Bn-1%7D%20" alt="LaTeX: \displaystyle \sum_{n=1}^{\infty}15\left( - \frac{3}{2} \right)^{n-1} " data-equation-content=" \displaystyle \sum_{n=1}^{\infty}15\left( - \frac{3}{2} \right)^{n-1} " /> . Since <img class="equation_image" title=" \displaystyle |r| = \frac{3}{2} \geq 1 " src="/equation_images/%20%5Cdisplaystyle%20%7Cr%7C%20%3D%20%5Cfrac%7B3%7D%7B2%7D%20%5Cgeq%201%20" alt="LaTeX: \displaystyle |r| = \frac{3}{2} \geq 1 " data-equation-content=" \displaystyle |r| = \frac{3}{2} \geq 1 " /> the geometric series diverges.</p> </p>