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1+3+....+103=?
Suma lui gaus nr impare


Răspuns :

(1+2+3+4+.....+103) - (2+4+6+8+....+102) =
[(103 · 104) : 2] - [2 · (1+2+3+....+51)] =
(103 · 52) - [2 · (51 · 52) : 2] =
5356 - (51 · 52) =
5356 - 2652 =
2704