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\large 
\underline
{MA2160 \hspace{5mm}Calculus II with Technology \hspace{50mm} Ch. 9-2, ex. 36}


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Class,  

Here is the correct solution to 36c

It is not possible to write the first 10 as 2*5 and write the sum from there as the rest of the terms have 2*10.  You must consider the sum from the second term on and add ten to the end or you can multiply the first 10 by 2 and subtract 10.  To include all bounces, I will do the second option.

\begin{tiny}
$$10+2(10(\frac{3}{4}))+2(10(\frac{3}{4})^2)+2(10(\frac{3}{4})^3)+...+2(10(\frac{3}{4})^n)=2*10+2(10(\frac{3}{4}))+2(10(\frac{3}{4})^2)+2(10(\frac{3}{4})^3)+...+2(10(\frac{3}{4})^n)-10$$
\end{tiny}


Given the right hand of the equation, we get:

$$\left( 2*\Sigma_{_{i=1}}^{^n}\,10\left( \frac{3}{4}\right)^{i-1} \right)-10$$
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