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Sa se calculeze suma primilor zece termeni ai progresiei geometrice in care b1=4 si q=1/2
Trebuie sa dea (2^10-1)/2^7
Cat mai clar va rog, de mult stau la acest exercitiu :(


Răspuns :

[tex]b_{1} = 4[/tex]

[tex]q = \frac{1}{2} [/tex]

[tex]Suma \: primilor \: n \: termeni \: (S_{n})[/tex]

[tex]S_{n}=b_{1}\times\frac{{q}^{n}-1}{q-1}[/tex]

[tex]S_{10}=4 \times \frac{ {( \frac{1}{2} )}^{10} - 1}{ \frac{1}{2} - 1} [/tex]

[tex]S_{10}=4 \times \frac{ \frac{ {1}^{10} }{ {2}^{10} } - 1}{ \frac{1}{2} - \frac{2}{2} } [/tex]

[tex]S_{10}=4 \times \frac{ \frac{1}{1024} - 1}{ \frac{1 - 2}{2} } [/tex]

[tex]S_{10}=4 \times \frac{ \frac{1}{1024} - \frac{1024}{1024} }{ - \frac{1}{2} } [/tex]

[tex]S_{10}=4 \times \frac{ \frac{1 - 1024}{1024} }{ - \frac{1}{2} } [/tex]

[tex]S_{10}=4 \times \frac{ - \frac{1023}{1024} }{ - \frac{1}{2} } [/tex]

[tex]S_{10}=4 \times \frac{1023}{1024} \times 2[/tex]

[tex]S_{10}= \frac{1023}{256} \times 2[/tex]

[tex]S_{10}= \frac{1023}{128} [/tex]