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The above is the answer in the notes but for the sum part I get $\frac{R(1-(1+\alpha))^{180}}{\alpha}$ instead?

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For a geometric sequence with first term equal to $a_1$ and ratio $q$, then the stands formula form the sum of the $n$ first elements is:

$$\frac{a_1.[q^{n}-1]}{q -1}=\frac{a_1.[1-q^{n}]}{1-q}$$

For this specific case we have a geometric sequence with first term equal to $1$ and ratio $1+\alpha$. Then we have that the sum of the $180$ first elements is:

$$\frac{1.[(1+\alpha)^{180}-1]}{(1+\alpha) -1}=\frac{(1+\alpha)^{180}-1}{\alpha}$$

Or

$$\frac{1.[1-(1+\alpha)^{180}]}{1-(1+\alpha)}=\frac{(1+\alpha)^{180}-1}{\alpha}$$

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    How did you get that? – 2017-01-15
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    I will write in more detail. – 2017-01-15
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    Please, take another look. Is that clear? – 2017-01-15
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    The sum of the first n is different on herehttp://mathworld.wolfram.com/GeometricSeries.html – 2017-01-15
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    It comes up as $1-r^{k+1}$ instead of $r^{k+1} - 1$ for the numerator – 2017-01-15
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    if you multiply by $-1$ the numerator and the denominator you will have the same formula. Take another look. – 2017-01-15
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    Ahhhhhhh I see, that makes a lot more sense thank you! – 2017-01-15
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    you are very welcome! – 2017-01-15