How do you find the sum of the finite geometric sequence of #Sigma 500(1.04)^n# from n=0 to 6?

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Answer 1 · 2017-12-16T16:53:10

See below.

The sum of a geometric series is given by:

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

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

From example:

#a=500# , #r=1,04# and #n=7# ( since we start at 0 )

#:.#

#500((1-(1.04)^7)/(1-1.04))~~3949.147#

#sum_(n=0)^(6)500(1.04)^n~~3949.147#