[pont494]

$2+2^2+2^3+...+2^n=2^{n+1}-2$

$1^2+3^2+5^2+...+(2n-1)^2=\frac{n}{3} (2n-1)(2n+1)$

$(1)(2)+(2)(3)+(3)(4)+...+n(n+1)=\frac{n}{3}(n+1)(n+2) $

$(1)(3)+(2)(4)+(3)(5)+...+n(n+2)=\frac{n}{6}(n+1)(2n+7) $

$\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +...+\frac{1}{2^n} = 1-\frac{1}{2^n} $