How do you find the Geometric formula for this sequence 1/3,1/9,1/27,1/81?

1 Answer
Jun 10, 2016

Answer:

Please see below.

Explanation:

A geometric series is written as #{a,ar,ar^2,ar^3,ar^4,.}#, where #a# is first term (of the series) and #r# is the ratio of a term to its preceding term.

#n^(th)# of such a series is #ar^(n-1)# and sum of the series #S_n# up to #n# terms is given by

#S_n=a(r^n-1)/(r-1)# (used if #|r|>1#)

or #S_n=a(1-r^n)/(1-r)# (used if #|r|<1#)

If #|r|<1#, sum of infinite series is #a/(1-r)#

Here #a=1/3# and #r=(1/9)/(1/3)=1/9xx3/1=1/3#

Hence, #n^(th)# term of the series is #1/3(1/3)^(n-1)=1/3^n# and sum of the series #S_n# is

#S_n=1/3((1-(1/3)^n))/((1-1/3))=1/3((1-(1/3^n)))/(2/3)#

= #1/3(1-(1/3^n))xx3/2#

= #1/2(1-(1/3^n)#

And sum of infinite series is #1/2#