How do you find the sum of the geometric series #1/16+1/4+1+...# to 7 terms?

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Answer 1 · 2016-11-12T23:57:18

#4^(-2)(1-4^7)/(1-4)=5461/16#

By collecting #4^(-2)# the progression can be rewritten as

#4^(-2)(1+4+4^2+....+4^6)#
in general we know that #1+q+q^2+....+q^n=(1-q^(n+1))/(1-q)#
in our case we have #4^(-2)(1-4^7)/(1-4)=5461/16#

Answer 2 · 2016-11-13T00:02:25

The sum of the first 7 terms is #5461/16= 341 5/16#

The formula for the sum of a particular number of terms of a geometric sequence is

#S_n=frac{a_1(1-r^n)}{1-r}# where

#a_1=#the first term of the sequence

#r=# the common ratio

To find #r#, divide a term by the previous term.

To find the sum of the first 7 terms of the sequence #1/16, 1/4, 1...#

#a_1=1/16#

#r=1/4 -: 1/16=4#

#n=7#

#S_n= frac{1/16(1-4^7)}{1-4} = 5461/16=341 5/16#