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Questions: Algebra BusinessCalculus
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Write the sum in sigma notation \(\displaystyle 13+\frac{143}{12}+\frac{1573}{144}\ldots\) and then find the sum.
Taking the ratio of the first two terms gives \(\displaystyle r = \frac{a_2}{a_1}=\frac{\frac{143}{12}}{13}=\frac{11}{12}\). Using \(\displaystyle a_{1}\) and \(\displaystyle r\) to write in sigma notation gives \(\displaystyle \sum_{n=1}^{\infty}13\left( \frac{11}{12} \right)^{n-1}\). Using the formula for the sum of an infinite geometric series gives \(\displaystyle \frac{13}{1-(\frac{11}{12})}=156\)
\begin{question}Write the sum in sigma notation $13+\frac{143}{12}+\frac{1573}{144}\ldots$ and then find the sum.
\soln{9cm}{Taking the ratio of the first two terms gives $r = \frac{a_2}{a_1}=\frac{\frac{143}{12}}{13}=\frac{11}{12}$. Using $a_{1}$ and $r$ to write in sigma notation gives $\sum_{n=1}^{\infty}13\left( \frac{11}{12} \right)^{n-1}$. Using the formula for the sum of an infinite geometric series gives $\frac{13}{1-(\frac{11}{12})}=156$}
\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 in sigma notation <img class="equation_image" title=" \displaystyle 13+\frac{143}{12}+\frac{1573}{144}\ldots " src="/equation_images/%20%5Cdisplaystyle%2013%2B%5Cfrac%7B143%7D%7B12%7D%2B%5Cfrac%7B1573%7D%7B144%7D%5Cldots%20" alt="LaTeX: \displaystyle 13+\frac{143}{12}+\frac{1573}{144}\ldots " data-equation-content=" \displaystyle 13+\frac{143}{12}+\frac{1573}{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{143}{12}}{13}=\frac{11}{12} " src="/equation_images/%20%5Cdisplaystyle%20r%20%3D%20%5Cfrac%7Ba_2%7D%7Ba_1%7D%3D%5Cfrac%7B%5Cfrac%7B143%7D%7B12%7D%7D%7B13%7D%3D%5Cfrac%7B11%7D%7B12%7D%20" alt="LaTeX: \displaystyle r = \frac{a_2}{a_1}=\frac{\frac{143}{12}}{13}=\frac{11}{12} " data-equation-content=" \displaystyle r = \frac{a_2}{a_1}=\frac{\frac{143}{12}}{13}=\frac{11}{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}13\left( \frac{11}{12} \right)^{n-1} " src="/equation_images/%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D13%5Cleft%28%20%5Cfrac%7B11%7D%7B12%7D%20%5Cright%29%5E%7Bn-1%7D%20" alt="LaTeX: \displaystyle \sum_{n=1}^{\infty}13\left( \frac{11}{12} \right)^{n-1} " data-equation-content=" \displaystyle \sum_{n=1}^{\infty}13\left( \frac{11}{12} \right)^{n-1} " /> . Using the formula for the sum of an infinite geometric series gives <img class="equation_image" title=" \displaystyle \frac{13}{1-(\frac{11}{12})}=156 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B13%7D%7B1-%28%5Cfrac%7B11%7D%7B12%7D%29%7D%3D156%20" alt="LaTeX: \displaystyle \frac{13}{1-(\frac{11}{12})}=156 " data-equation-content=" \displaystyle \frac{13}{1-(\frac{11}{12})}=156 " /> </p> </p>