Question

Vă rog să mă ajutati!!!!
Calculati:1/2-1/2^2-1/2^3-...-1/2^40
Si
1/2+1/2^2+1/2^3 +...+1/2^40 +1/2^40.
DAU COROANA!!!

Answer 1

1)
$S = \frac{1}{2} - \frac{1}{2^{2} } -\frac{1}{2^{3} }-...-\frac{1}{2^{40} } |\cdot 2 \Rightarrow 2S = 2\cdot( \frac{1}{2} - \frac{1}{2^{2} } -\frac{1}{2^{3} }-...-\frac{1}{2^{40} }) \Rightarrow \\ \\ \Rightarrow 2S = 1- \frac{1}{2} - \frac{1}{2^{2} } - ...- \frac{1}{2^{39} } -\frac{1}{2^{40}} } + \frac{1}{2^{40} }\Rightarrow$

$\Rightarrow 2S = 1- \frac{1}{2}- \frac{1}{2} + (\frac{1}{2}- \frac{1}{2^{2} } - ...- \frac{1}{2^{39} } -\frac{1}{2^{40}}) + \frac{1}{2^{40} } \Rightarrow \\ \\ 2S = 1- \frac{1}{2}- \frac{1}{2}+ S+ \frac{1}{2^{40} } \Rightarrow 2S-S = 1-1+ \frac{1}{2^{40} }\Rightarrow S = \frac{1}{2^{40} }$

2)
$S_{2} = \frac{1}{2} + \frac{1}{2^{2} } +\frac{1}{2^{3} }+...+\frac{1}{2^{40} }+\frac{1}{2^{40} } \Rightarrow \\ \\ \Rightarrow S_{2} = \frac{1}{2}+ \frac{1}{2} -(\frac{1}{2} - \frac{1}{2^{2} } -\frac{1}{2^{3} }-...-\frac{1}{2^{40} })+ \frac{1}{2^{40} }\Rightarrow \\ \\ \Rightarrow S_{2}= \frac{1}{2}+ \frac{1}{2} -S+ \frac{1}{2^{40} } \Rightarrow S_{2}=1- \frac{1}{2^{40} }+ \frac{1}{2^{40} } \Rightarrow S_{2} =1$