Евгений Колесников 11 months ago
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 </details> -->
 
-$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$
+<!-- $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ -->
 
 # Основы алгоримизации и программирования