Question

д поможіть будь ласка ​

Answer 1

Ответ:

a) $a_{n}$ = $(-3)^{n-1}$

$a_{1}$ = $(-3)^{0}$ = 1

$a_{2}$ = $(-3)^{1}$ = -3

$a_{3}$ = $(-3)^{2}$ = 9

$a_{4}$ = $(-3)^{3}$ = -27

б) $a_{n} = \frac{2^{n}}{3n}$

$a_{1} = \frac{2}{3}$

$a_{2} = \frac{2^{2}}{3*2} = \frac{4}{6} = \frac{2}{3}$

$a_{3} = \frac{2^{3}}{3*3} = \frac{8}{9}$

$a_{4} = \frac{2^{4}}{3*4} = \frac{16}{12} = \frac{4}{3} = 1\frac{1}{3}$

a) $a_{n} = (-2)^{n+1}$

$a_{1} = (-2)^{1+1} = (-2)^{2} = 4$

$a_{2} = (-2)^{3} = -8$

$a_{3} = (-2)^{4} = 16$

$a_{4} = (-2)^{5} = -32$

б) $a_{n} = \frac{2n}{3^{n}}$

$a_{1} = \frac{2}{3}$

$a_{2} = \frac{2*2}{3^{2}} = \frac{4}{9}$

$a_{3} = \frac{2*3}{3^{3}} = \frac{6}{27}$

$a_{4} = \frac{2*4}{3^{4}} = \frac{8}{81}$

Объяснение: