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Write the sum \(\displaystyle -42-41-40 \ldots -3-2\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=-41-(-42)=1\). Using the first term gives the sequene \(\displaystyle a_n= -42+(n-1)(1)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle -42+(n-1)(1)=-2 \implies n = 41 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{41} \left(n - 43\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 41(-42-2) }{2}=-902\).

Download \(\LaTeX\) 

\begin{question}Write the sum $-42-41-40 \ldots -3-2$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=-41-(-42)=1$. Using the first term gives the sequene $a_n= -42+(n-1)(1)$. Setting the general term equal to the last term and solving for $n$ gives $-42+(n-1)(1)=-2 \implies n = 41 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{41} \left(n - 43\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 41(-42-2) }{2}=-902$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Write the sum  <img class="equation_image" title=" \displaystyle -42-41-40 \ldots -3-2 " src="/equation_images/%20%5Cdisplaystyle%20-42-41-40%20%5Cldots%20-3-2%20" alt="LaTeX:  \displaystyle -42-41-40 \ldots -3-2 " data-equation-content=" \displaystyle -42-41-40 \ldots -3-2 " />  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-(-42)=1 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D-41-%28-42%29%3D1%20" alt="LaTeX:  \displaystyle a_2-a_1=-41-(-42)=1 " data-equation-content=" \displaystyle a_2-a_1=-41-(-42)=1 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= -42+(n-1)(1) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%20-42%2B%28n-1%29%281%29%20" alt="LaTeX:  \displaystyle a_n= -42+(n-1)(1) " data-equation-content=" \displaystyle a_n= -42+(n-1)(1) " /> . 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 -42+(n-1)(1)=-2 \implies n = 41  " src="/equation_images/%20%5Cdisplaystyle%20-42%2B%28n-1%29%281%29%3D-2%20%5Cimplies%20n%20%3D%2041%20%20" alt="LaTeX:  \displaystyle -42+(n-1)(1)=-2 \implies n = 41  " data-equation-content=" \displaystyle -42+(n-1)(1)=-2 \implies n = 41  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{41} \left(n - 43\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B41%7D%20%5Cleft%28n%20-%2043%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{41} \left(n - 43\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{41} \left(n - 43\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 41(-42-2) }{2}=-902 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2041%28-42-2%29%20%7D%7B2%7D%3D-902%20" alt="LaTeX:  \displaystyle \frac{ 41(-42-2) }{2}=-902 " data-equation-content=" \displaystyle \frac{ 41(-42-2) }{2}=-902 " /> . </p> </p>