What is the sum of the geometric sequence 1, 4, 16, … if there are 8 terms?

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Answer 1 · 2016-01-25T03:45:36

#a_2/a_1=4/1=4#

#a_3/a_2=16/4=4#

#implies# common ratio#=r=4# and #a_1=1#

Sum of geometric series is given by

#Sum=(a_1(1-r^n))/(1-r)#

Where #n# is number of terms, #a_1# is the first term, #r# is the common ratio.

Here #a_1=1#, #n=8# and #r=4#

#implies Sum=(1(1-4^8))/(1-4)=(1-65536)/-3=(-65535)/-3=21845#

Hence, the sum is #21845#