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Questions: Algebra BusinessCalculus
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Write the sum in sigma notation \(\displaystyle 7-\frac{91}{12}+\frac{1183}{144}\ldots\) and then find the sum.
Taking the ratio of the first two terms gives \(\displaystyle r = \frac{a_2}{a_1}=\frac{- \frac{91}{12}}{7}=- \frac{13}{12}\). Using \(\displaystyle a_{1}\) and \(\displaystyle r\) to write in sigma notation gives \(\displaystyle \sum_{n=1}^{\infty}7\left( - \frac{13}{12} \right)^{n-1}\). Since \(\displaystyle |r| = \frac{13}{12} \geq 1\) the geometric series diverges.
\begin{question}Write the sum in sigma notation $7-\frac{91}{12}+\frac{1183}{144}\ldots$ and then find the sum.
\soln{9cm}{Taking the ratio of the first two terms gives $r = \frac{a_2}{a_1}=\frac{- \frac{91}{12}}{7}=- \frac{13}{12}$. Using $a_{1}$ and $r$ to write in sigma notation gives $\sum_{n=1}^{\infty}7\left( - \frac{13}{12} \right)^{n-1}$. Since $|r| = \frac{13}{12} \geq 1$ the geometric series diverges.}
\end{question}
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\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Write the sum in sigma notation <img class="equation_image" title=" \displaystyle 7-\frac{91}{12}+\frac{1183}{144}\ldots " src="/equation_images/%20%5Cdisplaystyle%207-%5Cfrac%7B91%7D%7B12%7D%2B%5Cfrac%7B1183%7D%7B144%7D%5Cldots%20" alt="LaTeX: \displaystyle 7-\frac{91}{12}+\frac{1183}{144}\ldots " data-equation-content=" \displaystyle 7-\frac{91}{12}+\frac{1183}{144}\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{91}{12}}{7}=- \frac{13}{12} " src="/equation_images/%20%5Cdisplaystyle%20r%20%3D%20%5Cfrac%7Ba_2%7D%7Ba_1%7D%3D%5Cfrac%7B-%20%5Cfrac%7B91%7D%7B12%7D%7D%7B7%7D%3D-%20%5Cfrac%7B13%7D%7B12%7D%20" alt="LaTeX: \displaystyle r = \frac{a_2}{a_1}=\frac{- \frac{91}{12}}{7}=- \frac{13}{12} " data-equation-content=" \displaystyle r = \frac{a_2}{a_1}=\frac{- \frac{91}{12}}{7}=- \frac{13}{12} " /> . 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}7\left( - \frac{13}{12} \right)^{n-1} " src="/equation_images/%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D7%5Cleft%28%20-%20%5Cfrac%7B13%7D%7B12%7D%20%5Cright%29%5E%7Bn-1%7D%20" alt="LaTeX: \displaystyle \sum_{n=1}^{\infty}7\left( - \frac{13}{12} \right)^{n-1} " data-equation-content=" \displaystyle \sum_{n=1}^{\infty}7\left( - \frac{13}{12} \right)^{n-1} " /> . Since <img class="equation_image" title=" \displaystyle |r| = \frac{13}{12} \geq 1 " src="/equation_images/%20%5Cdisplaystyle%20%7Cr%7C%20%3D%20%5Cfrac%7B13%7D%7B12%7D%20%5Cgeq%201%20" alt="LaTeX: \displaystyle |r| = \frac{13}{12} \geq 1 " data-equation-content=" \displaystyle |r| = \frac{13}{12} \geq 1 " /> the geometric series diverges.</p> </p>