How do you find the sum of the finite geometric sequence of #Sigma 3(3/2)^n# from n=0 to 20?

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Answer 1 · 2018-03-28T00:41:18

#sum_(n=0)^(20)3(3/2)^n~~29921.31057#

We are asked to evaluate

#sum_(n=0)^(20)3(3/2)^n=3+3(3/2)+3(3/2)^2+3(3/2)^3+cdots+3(3/2)^20#

This is a geometric series of the form

#sum_(n=0)^kaR^n#

with #a=3# and #R=3/2#

There is a formula to calculate the #k#th partial sum of a geometric series. It is given by:

#S_k=a((1-R^(k+1))/(1-R))#

Therefore:

#S_20=a((1-R^21)/(1-R))=3((1-(3/2)^21)/(1-3/2))#

#=3((1-(3/2)^21)/(-1/2))=-6(1-(3/2)^21)=6((3/2)^21-1)#

Obviously this is going to be a pretty big number. If you punch it into a calculator, you'll find it's

#~~29921.31057#