How do you find the sum of the geometric series #Sigma 64(3/4)^(n-1)# n=1 to 8? Precalculus Series Sums of Geometric Sequences 1 Answer Hammer Jun 27, 2018 #S_8=(4^8-3^8)/4^7# Explanation: The sum of the first #n# terms of a geometric series with general term #b_m# and common ratio #q# is #S_n = b_1 * (q^n-1)/(q-1)# For #q<1#, this is usually written as #S_n = b_1 (1-q^n)/(1-q)# #b_n = 64(3/4)^(n-1)=> {(b_1 = 64),(q=b_(n+1)/b_n = 3/4) :}# #S_8 = 64*(1-(3/4)^8)/(1-3/4)=(4^8-3^8)/4^7# Answer link Related questions What is a sample problem about finding the sum of a geometric sequence? What is the formula for the sum of a geometric sequence? What is a sample problem about finding the sum of a geometric sequence? How do I find the sum of the geometric sequence #3/2#, #3/8#? What is the sum of the geometric sequence 3, 15, 75? What is the sum of the geometric sequence 8, 16, 32? How do I find the sum of the geometric series 8 + 4 + 2 + 1? How do you find the sum of the following infinite geometric series, if it exists. 2 + 1.5 +... How do you find the sum of the first 5 terms of the geometric series: 4+ 16 + 64…? How do you find S20 for the geometric series 4 + 12 + 36 + 108 + …? See all questions in Sums of Geometric Sequences Impact of this question 1679 views around the world You can reuse this answer Creative Commons License