\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus

Please login to create an exam or a quiz.

Algebra
Series
New Random

Write the sum \(\displaystyle 49+41+33 \ldots -335-343\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=41-(49)=-8\). Using the first term gives the sequene \(\displaystyle a_n= 49+(n-1)(-8)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle 49+(n-1)(-8)=-343 \implies n = 50 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{50} \left(57 - 8 n\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 50(49-343) }{2}=-7350\).

Download \(\LaTeX\) 

\begin{question}Write the sum $49+41+33 \ldots -335-343$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=41-(49)=-8$. Using the first term gives the sequene $a_n= 49+(n-1)(-8)$. Setting the general term equal to the last term and solving for $n$ gives $49+(n-1)(-8)=-343 \implies n = 50 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{50} \left(57 - 8 n\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 50(49-343) }{2}=-7350$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
\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}
HTML for Canvas 
<p> <p>Write the sum  <img class="equation_image" title=" \displaystyle 49+41+33 \ldots -335-343 " src="/equation_images/%20%5Cdisplaystyle%2049%2B41%2B33%20%5Cldots%20-335-343%20" alt="LaTeX:  \displaystyle 49+41+33 \ldots -335-343 " data-equation-content=" \displaystyle 49+41+33 \ldots -335-343 " />  in sigma notation and then find the sum.</p> </p>
HTML for Canvas
<p> <p>The common difference is given by  <img class="equation_image" title=" \displaystyle a_2-a_1=41-(49)=-8 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D41-%2849%29%3D-8%20" alt="LaTeX:  \displaystyle a_2-a_1=41-(49)=-8 " data-equation-content=" \displaystyle a_2-a_1=41-(49)=-8 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= 49+(n-1)(-8) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%2049%2B%28n-1%29%28-8%29%20" alt="LaTeX:  \displaystyle a_n= 49+(n-1)(-8) " data-equation-content=" \displaystyle a_n= 49+(n-1)(-8) " /> . Setting the general term equal to the last term and solving for  <img class="equation_image" title=" \displaystyle n " src="/equation_images/%20%5Cdisplaystyle%20n%20" alt="LaTeX:  \displaystyle n " data-equation-content=" \displaystyle n " />  gives  <img class="equation_image" title=" \displaystyle 49+(n-1)(-8)=-343 \implies n = 50  " src="/equation_images/%20%5Cdisplaystyle%2049%2B%28n-1%29%28-8%29%3D-343%20%5Cimplies%20n%20%3D%2050%20%20" alt="LaTeX:  \displaystyle 49+(n-1)(-8)=-343 \implies n = 50  " data-equation-content=" \displaystyle 49+(n-1)(-8)=-343 \implies n = 50  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{50} \left(57 - 8 n\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B50%7D%20%5Cleft%2857%20-%208%20n%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{50} \left(57 - 8 n\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{50} \left(57 - 8 n\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 50(49-343) }{2}=-7350 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2050%2849-343%29%20%7D%7B2%7D%3D-7350%20" alt="LaTeX:  \displaystyle \frac{ 50(49-343) }{2}=-7350 " data-equation-content=" \displaystyle \frac{ 50(49-343) }{2}=-7350 " /> . </p> </p>