Question

Oblicz sumę n początkowych ciągu geometrycznego

Answer 1

Odpowiedź:

a)

[tex]S_5 = a_1*\frac{1 - q^5}{1 - q} = 6*\frac{1 - (2/3)^5}{1 - 2/3} = 6*\frac{1 - 32/243}{1/3} =[/tex]  18 * [tex]\frac{211}{243} =[/tex] [tex]\frac{3786}{243} = \frac{422}{27 } = 15 \frac{17}{27}[/tex]

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b ) [tex]S_{21} = a_1*n = 18*21 = 378[/tex]

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c )   [tex]a_2 = 10[/tex]   [tex]a_4 = 20[/tex]      n = 5

[tex]a_4 = a_1*q^3 = a_1*q*q^2 = a_2*q^2 \\20 = 10*q^2[/tex]q² = 2

q = √2

[tex]a_2= a_1*q[/tex]

[tex]a_1 = 10 : \sqrt{2} = 5\sqrt{2}[/tex]

więc

[tex]S_{10} = a_1 *\frac{1 - q^{10}}{1 - q} = 5\sqrt{2} *\frac{1 - (\sqrt{2})^{10} }{1 - \sqrt{2} }[/tex]  [tex]= 5\sqrt{2} *\frac{1 - 2^5}{1 - \sqrt{2} } = 5\sqrt{2} *\frac{1 - 32}{1 - \sqrt{2} } = \frac{-155\sqrt{2} }{1 - \sqrt{2} } *\frac{1 + \sqrt{2} }{1 + \sqrt{2} } =[/tex]  [tex]\frac{- 155\sqrt{2} -310}{1 - 2} = 155\sqrt{2} + 310[/tex]

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d )   [tex]a_3 = 6[/tex]       [tex]a_6 = - \frac{1}{36}[/tex]           n = 5

[tex]a_6 = a_3*q^3\\q^3 = a_6 : a_3 = - \frac{1}{36} : 6 = - \frac{1}{216}[/tex]

q = - [tex]\frac{1}{6}[/tex]

[tex]a_1 = a_3 : q^2 = 6 : \frac{1}{36} = 216[/tex]

[tex]S_5 = 216*\frac{1 - q^5}{1 - q} =[/tex]  [tex]\frac{216 - 216*q^5}{1 - q} =[/tex][tex]\frac{216 +\frac{1}{36} }{\frac{7}{6} } = \frac{6}{7} *( 216 \frac{1}{36} )[/tex] [tex]= \frac{6}{7} *\frac{7777}{36} = \frac{1111}{6} = 185 \frac{1}{6}[/tex]

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