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Questions: Algebra BusinessCalculus
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Write the sum \(\displaystyle -1-3-5 \ldots -97-99\) in sigma notation and then find the sum.
The common difference is given by \(\displaystyle a_2-a_1=-3-(-1)=-2\). Using the first term gives the sequene \(\displaystyle a_n= -1+(n-1)(-2)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle -1+(n-1)(-2)=-99 \implies n = 50 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{50} \left(1 - 2 n\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 50(-1-99) }{2}=-2500\).
\begin{question}Write the sum $-1-3-5 \ldots -97-99$ in sigma notation and then find the sum.
\soln{9cm}{The common difference is given by $a_2-a_1=-3-(-1)=-2$. Using the first term gives the sequene $a_n= -1+(n-1)(-2)$. Setting the general term equal to the last term and solving for $n$ gives $-1+(n-1)(-2)=-99 \implies n = 50 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{50} \left(1 - 2 n\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 50(-1-99) }{2}=-2500$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Write the sum <img class="equation_image" title=" \displaystyle -1-3-5 \ldots -97-99 " src="/equation_images/%20%5Cdisplaystyle%20-1-3-5%20%5Cldots%20-97-99%20" alt="LaTeX: \displaystyle -1-3-5 \ldots -97-99 " data-equation-content=" \displaystyle -1-3-5 \ldots -97-99 " /> in sigma notation and then find the sum.</p> </p>
<p> <p>The common difference is given by <img class="equation_image" title=" \displaystyle a_2-a_1=-3-(-1)=-2 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D-3-%28-1%29%3D-2%20" alt="LaTeX: \displaystyle a_2-a_1=-3-(-1)=-2 " data-equation-content=" \displaystyle a_2-a_1=-3-(-1)=-2 " /> . Using the first term gives the sequene <img class="equation_image" title=" \displaystyle a_n= -1+(n-1)(-2) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%20-1%2B%28n-1%29%28-2%29%20" alt="LaTeX: \displaystyle a_n= -1+(n-1)(-2) " data-equation-content=" \displaystyle a_n= -1+(n-1)(-2) " /> . 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 -1+(n-1)(-2)=-99 \implies n = 50 " src="/equation_images/%20%5Cdisplaystyle%20-1%2B%28n-1%29%28-2%29%3D-99%20%5Cimplies%20n%20%3D%2050%20%20" alt="LaTeX: \displaystyle -1+(n-1)(-2)=-99 \implies n = 50 " data-equation-content=" \displaystyle -1+(n-1)(-2)=-99 \implies n = 50 " /> . Writing in sigma notation gives <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{50} \left(1 - 2 n\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B50%7D%20%5Cleft%281%20-%202%20n%5Cright%29%20" alt="LaTeX: \displaystyle \displaystyle \sum_{n=1}^{50} \left(1 - 2 n\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{50} \left(1 - 2 n\right) " /> . Using the formula for a finite arithmetic sum gives <img class="equation_image" title=" \displaystyle \frac{ 50(-1-99) }{2}=-2500 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2050%28-1-99%29%20%7D%7B2%7D%3D-2500%20" alt="LaTeX: \displaystyle \frac{ 50(-1-99) }{2}=-2500 " data-equation-content=" \displaystyle \frac{ 50(-1-99) }{2}=-2500 " /> . </p> </p>