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On average, when a basketball is dropped, it bounces 7/10 as high as it did on the previous bounce. If a ball is dropped from a height of 10 m, determine the total distance up and down that the ball travels by top of the 11th bounce. Round to the nearest hundredth

Mathematics
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\[10(7\div10)^{11}\]
it is the sum of a geometric sequence Sum = 10[(1-.7^11)/(1-.7)] = 32.67
Are you familiar with the sum of a geometric series? Let \[S = 1 + x + x^2 + x^3 +...+x^n\] then \[xS = x + x^2 + x^3 + ...+x^{n+1}\] and so \[S - xS = S(1-x) = 1 - x^{n+1}\] So \[S = \frac{1-x^{n+1}}{1-x}\] That's the sum of a geometric series. Now examining your problem, we have to add the following: \[ 10 + \frac{7}{10} \cdot 10 + \frac{7}{10}\cdot 10 + \left(\frac{7}{10}\right)^2 \cdot 10 + \left(\frac{7}{10}\right)^2 \cdot 10 + ... \] so on and so forth. Noting that the distance traveled (one-way, either up or down) after the nth bounce is \[\left(\frac{7}{10}\right)^n \cdot 10\] and that everything except the first and last bounce is repeated, we can say that our total distance traveled is \[\left[2\sum_{n=0}^{11} \left(\frac{7}{10}\right)^n\cdot 10\right]- 10 - \left(\frac{7}{10}\right)^{11}\cdot 10\] but this looks just like a geometric series with x = 7/10, so that means \[\sum_{n=0}^{11} \left(\frac{7}{10}\right)^n = \frac{1-\left(\frac{7}{10}\right)^{12}}{1-\frac{7}{10}}\] And the rest is just calculation... :)

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