Question

Help with Geometric Series Sigma problem?

Answer 1

\[\sum_{n=1}^{6}(1/2)^{n-1}\] I know that the answer is 63/2 but I don't know how to get it

Answer 2

oops i mean 63/32

Answer 3

use this formula

\(\huge S_n = \frac{a(r^n -1)}{r-1}\)

where
a is your first term which is 1,
r is common ratio => which is 1/2
n is your no of terms.

Answer 4

well i know the other formulas we are supposed to use are:
\[\S _{n}=a _{1}-a _{1}r ^{n}/1-r\]
and
\[\S _{n}=a _{1}-a _{1}r/1-r\]
but I don't understand how you find \[a _{1}\] or r or n

Answer 5

your series

\[ \sum_{i=1}^{6}(1/2)^{n-1} = 1*(1/2)^0 + 1*(1/2)+1*(1/2)^2+ ...+1*(1/2)^5\]

from here, you have six terms, n=6, you can count them.
your a, is first term, which is 1
and your r is (1/2)

Answer 6

how do you know what that series is? I don't really understand sigmas altogether... or what the different parts of it are (like what you do with the 6 and the n=1 and the (1/2)^n-1 )

Answer 7

put the values of n in 1/2^(n-1) and increase n keep adding then until n=6

Answer 8

what does the n=1 mean on the bottom of the sigma? when you insert it into the 1/2^n-1 is that your first number that starts out the series? and the number on top is the last number of the series?

Answer 9

yes ,,, and it goes untill it reaches top value

Answer 10

ok so thats how u figure out what \[a_{1}\] is right? so then how do you figure out what n and r are?