#a_2/a_1=8/2=4#
#a_3/a_2=32/8=4#
#implies# common ration #=r=4# and the series is geometric.
Nth term of a geometric series is given by
#T_n=a_1r^(n-1)#
Where #T_n# is the nth term, #a_1# is the first term, #r# is the common ration and #n# is the number of terms.
Here #a_1=2#, #r=4# Last term#=131072#
Let last term be the nth term
#implies 131072=2*(4)^(n-1)#
#implies 4^(n-1)=65536#
#implies 4^(n-1)=4^8#
#implies (n-1)=8#
#implies n=9#
Sum of geometric series is given by
#Sum=(a_1(1-r^n))/(1-r)#
Where #r# is the common ratio #a_1# is the first term and #n# is the number of terms.
#implies Sum=(2(1-4^9))/(1-4)=(2(1-262144))/-3=(2(-262143))/-3=(-524286)/-3=174762#
#implies Sum=174762#.
