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# Series Question watch

1. $\frac{1}{2} + \frac{1}{6} + \frac{1}{18} + ...$

Find the least number of terms of the series which must be taken for their sum to exceed 2999/4000.
2. So you know it's a geometric series, since your dividing by 3 each time. Just plug in the forumla for the sum of the terms of a gemotric sequence and plug in the numbers for the sum being greater than (turn it into an inequality) 2999/4000, a = 1/2, r= 1/3,
3. it is a G.P. so the sum is

a( 1 - rn )/ ( 1 - r )

and you want this to be > 2999/4000

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