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Math Help - Geometric series....

  1. #1
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    Geometric series....

    I'm trying to solve the following problem: Suppose you bought a 56 cm houseplant. During the first year, the plant grows 24cm. Each following year, it grows 3/4 of the previous year's growth; meaning it grows 18cm during the second year, 13.5cm during the third year, and so on. Now, what would be the height of the tree at the end of the 20th year? At the end of the 30th year? If it could grew forever following the same patter, how tall would it grow?

    My attempt:

    S_{n} = a\left(\frac{1-r^n}{1-r}\right)

    S_{20} = 24\left(\frac{1-\left(\frac{3}{4}\right)^{20}}{1-\frac{3}{4}}\right)  = 95.69556365


     S_{20}+56 = 24\left(\frac{1-\left(\frac{3}{4}\right)^{20}}{1-\frac{3}{4}}\right) + 56  = 95.69556365+56 = \boxed{151.6955637}<br />

    S_{30} = 24\left(\frac{1-\left(\frac{3}{4}\right)^{30}}{1-\frac{3}{4}}\right)  = 95.98285612

    S_{30}+56 = 24\left(\frac{1-\left(\frac{3}{4}\right)^{30}}{1-\frac{3}{4}}\right)+56  = 95.98285612+56 = \boxed{151.9828561}

    S_{\infty} = \frac{1}{1-r} = \frac{24}{1-\left(\frac{3}{4}\right)} = \boxed{96}<br />

    S_{\infty}+56 = \frac{1}{1-r}+56 = \frac{24}{1-\left(\frac{3}{4}\right)}+56 = 96+56 = \boxed{152}<br />

    Hope it's right.
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  2. #2
    Senior Member apcalculus's Avatar
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    Hi Mozart, I don't see any problems with your solution.
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