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# The annual profit for the H. J. Heinz Company from 1980 through 1989 can be approximated by the model $a_n=167.5 e^{0.12 n}$, n=0,1,2, ..., 9 where $a_n$ is the annual profit in millions of dollars and n represents the year, with n=0 corresponding to 1980 . Use the formula for the sum of a geometric sequence to approximate the total profit earned during this 10-year period.

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We want to approximate the profit earned during the 10 years, which is given by $a_0 + a_1 + \cdot + a_9$, and since we know that $a_n = 167.5 \cdot e^{0.12 \cdot n}$, we have:

\begin{align*} \text{profit } &= \sum_{n = 0}^{9} a_n \\ &= \sum_{n = 0}^9 167.5 \cdot e^{0.12 \cdot n} \\ &= \sum_{n = 0}^9 167.5 \cdot \left( e^{0.12} \right)^n \\ &= \dfrac{167.5 \cdot \left(1 - \left( e^{0.12} \right)^{10}\right)}{1 - e^{0.12} } \\ &\approx \color{#4257b2} 3048.07 \end{align*}

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