What is the sum of an infinite geometric series with a first term of 6 and a common ratio of 1/2?

1 Answer
George C.
Feb 9, 2017

#12#

Explanation:

The general term of a geometric series can be represented by the formula:

#a_n = a r^(n-1)#

where #a# is the initial term and #r# the common ratio.

We find:

#(1-r) sum_(n=1)^N ar^(n-1) = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = sum_(n=1)^N ar^(n-1) - sum_(n=2)^(N+1) ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a + color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - ar^N#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a(1 - r^N)#

Dividing both ends by #(1-r)# we find:

#sum_(n=1)^N ar^(n-1) = (a(1 - r^N))/(1-r)#

If #abs(r) < 1# then #lim_(N->oo) r^N = 0# and we find:

#sum_(n=1)^oo ar^(n-1) = lim_(N->oo) (a(1 - r^N))/(1-r) = a/(1-r)#

In the given example, #a=6# and #r=1/2#. So #abs(r) < 1# and:

#sum_(n=1)^oo 6*(1/2)^(n-1) = 6/(1-1/2) = 12#

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