Question

Salut, am nevoie de ajutor la ex. 2.559A

Answer 1

2⁰+2⁻¹+2⁻²+2⁻³+...+2⁻¹⁰ =

(b₁ = 2⁰, q = 2⁻¹, n = 11)

= 2⁰•[(2⁻¹)¹¹-1]/(2⁻¹-1) = (2⁻¹¹-1)/(2⁻¹-1) =

= (2⁻¹¹-1)/(-2⁻¹) = (1-2⁻¹¹)/(2⁻¹) =

= (2¹¹-1)/(2¹⁰) = (2¹²-2)/(2¹¹)

⇒ c) corect.

Sau așa:

$S = 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}} \\\\\dfrac{S}{2} = \quad\,\,\,\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}+\dfrac{1}{2^{11}} \\\\\noindent\rule{5.4cm}{0.7pt} \\\\ \dfrac{S}{2}-S = \dfrac{1}{2^{11}}-1 \\ \\ -\dfrac{S}{2} = \dfrac{1-2^{11}}{2^{11}} \\ \\\\ \Rightarrow \boxed{S = \dfrac{2^{12}-2}{2^{11}}}$