Hallo be,
\(\sum\limits_{i=0}^{n} \) qn = (qn - 1) / (q - 1)
\(\sum\limits_{i=99}^{999} \) 3i-1 = \(\sum\limits_{i=98}^{998}\) 3i = \(\sum\limits_{i=0}^{998} \) 3i - \(\sum\limits_{i=0}^{97} \) 3i
= (3998 - 1) / (3-1) - (397 - 1) / 3 - 1)
≈ 7,344837886 ·10475 [ sagt mein Computer :-) ]
Gruß Wolfgang