SOLUTION: The sum of a finite geometric series.
Given n=6
-5, -15, -45, ...,
I worked this out and ended up with -1820 , but I am not sure the result is correct.
a1(1−r^n)
Algebra.Com
Question 1109416: The sum of a finite geometric series.
Given n=6
-5, -15, -45, ...,
I worked this out and ended up with -1820 , but I am not sure the result is correct.
a1(1−r^n)/1−r
-5(1-3^6/1-2 = 3640/-2 = -1820
Please let me know if the formula I used was incorrect, thanks!
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Can check the result
-5-15-45-135-405-1215=-1820
you are correct, except the denominator is 1-3 to get -2, not 1-2.
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