Geometric Series Review

a_n=a₁(r)^n-1

137,781=7(3)^n-1

19683=3^n-1

Guess and check the answer OR use logs

$$\frac{\log\left(19683\right)}{\log\left(3\right)}=9\ =\ n-1\$$  

an=a1(r)n-1

62.5 = 8000(1/2)n-1

1/128 = (1/2)n-1

Guess and check or use LOGS

$$\frac{\log\left(\frac{1}{128}\right)}{\log\left(\frac{1}{2}\right)}=7\ =\ n-1$$  

n=8

$$S_n=a_1\ \frac{1-r^n}{1-r}$$  

$$S_8=8000\ \frac{1-\left(\frac{1}{2}\right)^8}{1-\left(\frac{1}{2}\right)}=15937.5$$  

$$S_n=a_1\left(\frac{1-r^n}{1-r}\right)$$  

$$S_5=-2\left(\frac{1-4^5}{1-4}\right)$$  

a₄/a₁ = 64

since there are three "steps" between a₁ and a₄: r³ = 64

$$r=\sqrt[3]{64}$$  

r=4

an=a1(r)n-1

6.25=800(1/2)n-1

1/128 = (1/2)n-1

Guess and check or use LOGS

$$\frac{\log\ \left(\frac{1}{128}\right)}{\log\left(\frac{1}{2}\right)}=7=n-1$$  

n=8

$$S_n=a_1\left(\frac{1-r^n}{1-r}\right)$$

$$800\left(\frac{1-\left(\frac{1}{2}\right)^8}{1-\left(\frac{1}{2}\right)}\right)$$  

$$S_n=a_1\left(\frac{1-r^n}{1-r}\right)$$

$$S_8=1\left(\frac{1-5^8}{1-5}\right)$$  

$$S_n=a_1\left(\frac{1-r^n}{1-r}\right)$$

$$S_{12}=6\left(\frac{1-\left(-2\right)^{12}}{1-\left(-2\right)}\right)$$  

if you got 8194 you forgot to put the -2 in parenthesis

$$S_n=a_1\left(\frac{1-r^n}{1-r}\right)$$

$$S_0=-3\left(\frac{1-3^9}{1-3}\right)$$  

$$S_n=a_1\left(\frac{1-r^n}{1-r}\right)$$

$$S_8=-4\left(\frac{1-\left(-4\right)^8}{1-\left(-4\right)}\right)$$  

$$S=\frac{a_1}{1-r}$$  

$$S=\frac{4}{1-\left(-\frac{1}{3}\right)}$$