Question

Calculati S=a^0 + a^1 + a^2 +. + a^2011, stiind ca a=2. Multumesc!.

Answer 1

Explicație pas cu pas:

progresie geometrică:

$b_{1} = 2^{0} = 1 \ ; \ q = 2 \ ; \ n = 2012$

$S_{n} = \dfrac{b_{1}\cdot ({q}^{n} - 1) }{q - 1}$

$S_{2012} = {2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{2011} =$

$= \dfrac{1\cdot ({2}^{2012} - 1) }{2 - 1} = {2}^{2012} - 1$