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How do you find the sum of the geometric series #Sigma 64(3/4)^(n-1)# n=1 to 8?

1 Answer

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#

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