Solutions - 0150-03

(a)

First term:

$$a_{-4}={(-\frac{4}{9})}^{-4}\gg \gg \frac{3^8}{2^8}$$

Common ratio is $r=-4/9$.

Geometric series summation formula, always first term on top:

$$\begin{array}{c}S=\frac{a_{-4}}{1-r}\gg \gg \frac{3^8}{2^8}\left(\frac{1}{1--4/9}\right)\\ \\ \gg \gg \frac{3^8}{2^8}\frac{9}{13}\gg \gg \frac{3^{10}}{2^8(13)}\gg \gg \approx 17.74\end{array}$$

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(b) ${\sum }_{n=0}^{\infty }{e}^{3-2n}$

First term: $a_0=e^3$

Common ratio:

$$r=\frac{e^{3-2(n+1)}}{e^{3-2n}}\gg \gg \frac{e\cdot e^{-2n}}{e^3\cdot e^{-2n}}\gg \gg \frac{1}{e^2}$$

Geometric series summation formula:

$$S=\frac{a_0}{1-r}\gg \gg \frac{e^3}{1-\frac{1}{e^2}}\gg \gg \frac{e^5}{e^2-1}$$