# can someone check my work? geometric series

• Oct 23rd 2006, 08:15 AM
frogsrcool
can someone check my work? geometric series
okay I attached the two questions and I will show my work here
question#1
sn= a(1-r^n) / (1-r)

a = 16
r = 0.5
n = 50
so:
sn= a(1-r^n)/ (1-r)
sn= 16(1-(0.5)^50)/ (1-(0.5))
sn= 16(1-(0.5)^50)/ (0.5)
sn= 32

question#2
sn= a(1-r^n)/ (1-r)
a = 3*(1-.07)
r =(1-0.07)
n = 7
So..
sn= a(1-r^n)/ (1-r)
sn= 3(1.07)x(1-(1-0.07)^7)/ (1-(1-0.07))
sn= 3(1.07)x (1-(1-0.07)^7)/(.07)
sn= 1.83 I rounded to two decimal places!

thanks for checking!
• Oct 23rd 2006, 08:44 AM
topsquark
Quote:

Originally Posted by frogsrcool
okay I attached the two questions and I will show my work here
question#1
sn= a(1-r^n) / (1-r)

a = 16
r = 0.5
n = 50
so:
sn= a(1-r^n)/ (1-r)
sn= 16(1-(0.5)^50)/ (1-(0.5))
sn= 16(1-(0.5)^50)/ (0.5)
sn= 32

question#2
sn= a(1-r^n)/ (1-r)
a = 3*(1-.07)
r =(1-0.07)
n = 7
So..
sn= a(1-r^n)/ (1-r)
sn= 3(1.07)x(1-(1-0.07)^7)/ (1-(1-0.07))
sn= 3(1.07)x (1-(1-0.07)^7)/(.07)
sn= 1.83 I rounded to two decimal places!

thanks for checking!

The first one is correct. The setup for the second one is correct, but you have a math error somewhere. I can point out one thing: Your a = 1-0.07 = 0.93, not 1.07. Anyway, I got the sum to be about 15.8750653.

-Dan
• Oct 23rd 2006, 08:58 AM
Soroban
Hello, frogsrcool!

Quote:

$\displaystyle 1)\;\sum^{50}_{n=1} 8(0.5)^{n-2}$

$\displaystyle S_n\:= \:a\frac{1 - r^n}{1 - r}$

$\displaystyle a = 16,\;r = 0.5,\;n = 50$

so: .$\displaystyle S_{50}\:=\:16\frac{1 - 0.5^{50}}{1 - 0.5}$

Therefore: .$\displaystyle S_{50} \:\approx\:32$

Since $\displaystyle (0.5)^{50} \:\approx\:8.88 \times 10^{-16}$, your answer is close enough!

Quote:

$\displaystyle 2)\;\sum^7_{n=1}3(1-0.07)^n$

$\displaystyle a \;= \;3(1 - 0.07) \;=\;3(0.93) \;=\;2.79$
$\displaystyle r \;= \; 1 - 0.07 \;=\;0.93$
$\displaystyle n \;= \;7$

$\displaystyle S_7 \;=\;2.79\,\frac{1 - 0.93^7}{1 - 0.93} \;=\;2.79\,\frac{0.398299129}{0.07} \;=\;15.8750653$