Question

вычислите: (2^10/3)+(2^9/3)+(2^8/3)+...+(2^2/3)+(2/3)+1/3.

Answer 1

$\frac{ 2^{10} }{3} + \frac{2^9}{3} + \frac{2^8}{3}+...+ \frac{2^2}{3}+ \frac{2^1}{3}+ \frac{2^0}{3}= \\ \\ \frac{1}{3} *( 2^{10} +2^9+2^8+...+2^2+2^1+2^0) \\ \\ b_{1} = 2^{10} \\ \\ q= \frac{1}{2} \\ \\ S_{n} = \frac{ b_{1} (1-q^n)}{1-q} \\ \\ \frac{1}{3} * \frac{2^{10} *(1-( \frac{1}{2})^{11}) }{1- \frac{1}{2} } =\frac{1}{3} *( 2^{10} *2* \frac{ 2^{11}-1 }{ 2^{11} })= \\ \\ =\frac{1}{3} *( 2^{11} -1)=\frac{1}{3} *(2048-1)= \frac{2047}{3} =682 \frac{1}{3}$