SOLUTION: sum of series {{{1/sqrt(3) + 1 + 3/sqrt(3)+.... to 18}}} terms is
a.{{{9841(1+sqrt(3))/sqrt(3)}}}
b.9841
c.9841/root3
d.none of these
Algebra.Com
Question 878674: sum of series terms is
a.
b.9841
c.9841/root3
d.none of these
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
S=(1/sqrt(3))*(1 - sqrt(3)^18)/(1 - sqrt(3))
a.9841(1+root3)/root3
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