Question

Сумма ряда 1+1/3+1/9+1\27+.... равна

Answer 1

Ответ:

$S_{0} = 0$

Пошаговое объяснение:

$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$

$a_{n} = a_{1} r^{n-1}$

$a_{n} = (1) * (\frac{1}{3} )^{n-1}$

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

$a_{n} = \frac{1^{n-1} }{3^{n-1} }$

$a_{n} = \frac{1}{3^{n-1} }$

$S_{n} = \frac{a_{1}(r^{n}- 1) }{r - 1}$

$S_{0} = 1 * \frac{(\frac{1}{3})^{0}- 1 }{\frac{1}{3}- 1 }$

$S_{0} = \frac{(\frac{1}{3})^{0}- 1 }{\frac{1}{3}- 1 }$

$S_{0} \frac{3((\frac{1}{3})^{0}- 1) }{3(\frac{1}{3}- 1) }$

$S_{0} =\frac{3(\frac{1}{3})^{0}+ 3 * -1 }{3(\frac{1}{3})+3* -1 }$

$S_{0} = \frac{3(\frac{1}{3})^{0}+ 3 * -1 }{1+3*-1}$

$S_{0} = \frac{0}{1+3*-1}$

$S_{0} = \frac{0}{-2}$

Ответ: $S_{0} = 0$