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CIĄGI GEOMETRYCZNE NA JUZ DAJE NAJ ITP​



CIĄGI GEOMETRYCZNE NA JUZ DAJE NAJ ITP class=

Odpowiedź :

Basetla

[tex]a_{n} = \frac{2}{5^{n+2}}\\\\a_{n+1} = \frac{2}{5^{n+1+2}} = \frac{2}{5^{n+3}}\\\\q = \frac{a_{n+1}}{a_{n}} = \frac{\frac{2}{5^{n+3}}}{\frac{2}{5^{n+2}}}=\frac{5^{n+2}}{5^{n+3}} = 5^{n+2-(n+3)} = 5^{n+2-n-3} = 5^{-1} = \frac{1}{5}\\\\q = \frac{1}{5} \ \ \Rightarrow \ \ ciag \ \{a_{n}\} \ jest \ geometryczny[/tex]

Catta1eya

[tex]a_n = \frac{2}{5^{n+2}}\\\\a_1=\frac{2}{5^{1+2}} = \frac{2}{5^3} = \frac{2}{125}\\a_2=\frac{2}{5^{2+2}} = \frac{2}{5^4} = \frac{2}{625}\\a_3 = \frac{2}{5^{3+2}} = \frac{2}{5^5} = \frac{2}{3125}[/tex]

[tex]\frac{a_2}{a_1} = \frac{2}{625} * \frac{125}{2} = \frac{125}{625}=\frac{5}{25} = \frac{1}{5}\\\frac{a_3}{a_2} = \frac{2}{3125} * \frac{625}{2} = \frac{625}{3125} = \frac{1}{5}[/tex]

Odp. Tak, ciąg jest geometryczny, iloraz jest stały i wynosi [tex]\frac{1}{5}[/tex]

Wzór ogólny:

[tex]a_n = \frac{2}{125} * (\frac{1}{5})^{n-1}[/tex]

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