How do you find the sum of #8(3/4)^(i-1)# from i=1 to 10?

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Answer 1 · 2017-10-05T11:49:37

#sum^10 8 (3/4)^(i-1) ~~ 30.198#
#i=1#

#sum^10 8 (3/4)^(i-1) = ?# .
#i=1#
The 1st term #(i=1)# is #8 (3/4)^(i-1) = 8 (3/4)^(1-1)=8*1=8#

2nd term #(i=2)# is #8 (3/4)^(i-1) = 8 (3/4)^(2-1)=8*3/4#

3rd term #(i=3)# is #8 (3/4)^(i-1) = 8 (3/4)^(3-1)=8*(3/4)^2#

4th term #(i=4)# is #8 (3/4)^(i-1) = 8 (3/4)^(4-1)=8*(3/4)^3#

The series is # 8, 8*3/4, 8*(3/4)^2 , 8*(3/4)^3 ......,8*(3/4)^9#.

This is a geometric sequence with first term is #a_1=8# and

common ratio is #r=3/4=0.75# and number of terms is #n=10#

Sum formula of geometric sequence is #S_n=a_1((r^n-1)/(r-1))#

#:. S_10=8((0.75^10-1)/(0.75-1)) ~~ 30.198 :.#

#sum^10 8 (3/4)^(i-1) ~~ 30.198# [Ans]
#i=1#