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Write the sum \(\displaystyle 41+45+49 \ldots +189+193\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=45-(41)=4\). Using the first term gives the sequene \(\displaystyle a_n= 41+(n-1)(4)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle 41+(n-1)(4)=193 \implies n = 39 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{39} \left(4 n + 37\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 39(41+193) }{2}=4563\).

Download \(\LaTeX\) 

\begin{question}Write the sum $41+45+49 \ldots +189+193$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=45-(41)=4$. Using the first term gives the sequene $a_n= 41+(n-1)(4)$. Setting the general term equal to the last term and solving for $n$ gives $41+(n-1)(4)=193 \implies n = 39 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{39} \left(4 n + 37\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 39(41+193) }{2}=4563$. }

\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 41+45+49 \ldots +189+193 " src="/equation_images/%20%5Cdisplaystyle%2041%2B45%2B49%20%5Cldots%20%2B189%2B193%20" alt="LaTeX:  \displaystyle 41+45+49 \ldots +189+193 " data-equation-content=" \displaystyle 41+45+49 \ldots +189+193 " />  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=45-(41)=4 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D45-%2841%29%3D4%20" alt="LaTeX:  \displaystyle a_2-a_1=45-(41)=4 " data-equation-content=" \displaystyle a_2-a_1=45-(41)=4 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= 41+(n-1)(4) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%2041%2B%28n-1%29%284%29%20" alt="LaTeX:  \displaystyle a_n= 41+(n-1)(4) " data-equation-content=" \displaystyle a_n= 41+(n-1)(4) " /> . 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 41+(n-1)(4)=193 \implies n = 39  " src="/equation_images/%20%5Cdisplaystyle%2041%2B%28n-1%29%284%29%3D193%20%5Cimplies%20n%20%3D%2039%20%20" alt="LaTeX:  \displaystyle 41+(n-1)(4)=193 \implies n = 39  " data-equation-content=" \displaystyle 41+(n-1)(4)=193 \implies n = 39  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{39} \left(4 n + 37\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B39%7D%20%5Cleft%284%20n%20%2B%2037%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{39} \left(4 n + 37\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{39} \left(4 n + 37\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 39(41+193) }{2}=4563 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2039%2841%2B193%29%20%7D%7B2%7D%3D4563%20" alt="LaTeX:  \displaystyle \frac{ 39(41+193) }{2}=4563 " data-equation-content=" \displaystyle \frac{ 39(41+193) }{2}=4563 " /> . </p> </p>