The infinite sum is (first term) divided by (1 minus the common ratio). The first term is 6(1/4) = 3/2; the common ratio is 1/4:
[tex]S=\frac{\frac{3}{2} }{1-\frac{1}{4} } =\frac{\frac{3}{2} }{\frac{3}{4} } =(\frac{3}{2} )*(\frac{4}{3} )=2[/tex]
The terms after n=16 are tiny and will certainly bring the sum down only a tiny bit.
So to the nearest whole number the sum of the finite geometric series is 2.
