Let
denote the sum of the first
terms of a geometric progression.
Consider that and What is the greatest possible integer value for ?
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We have that ∑ 2 n − ∑ n ∑ n = ∑ 3 n − ∑ 2 n ∑ 2 n − ∑ n . (proof left to the reader).
Substituting, we find x − 5 0 5 0 = 6 5 0 − x x − 5 0 ⇔ x 2 − 5 0 x − 3 0 0 0 0 = 0 → x 1 = − 1 5 0 , x 2 = 2 0 0 .