How do you find the sum of the finite geometric sequence of #Sigma 5(3/5)^n# from n=0 to 40?

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Answer 1 · 2017-10-13T20:00:31

12.5 (1.d.p)

First calculate the first three terms:

#5(3/5)^0= color(blue)(5) , 5(3/5)^1=color(blue)(3) , 5(3/5)^2=color(blue)(9/5)#

Find the common ratio:

#3/5 = (9/5)/3=3/5#

The sum of a geometric sequence is:

#a((1-r^n)/(1-r))#

Where a is the first term, n is the nth term and r is the common ratio.

So:

#5((1-(3/5)^40)/(1-(3/5)))=12.4999999899743791#(16 .d.p.)