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terrellmakinmovessmi
10.07.2019 •
Mathematics
Identify whether the series summation of 12 open parentheses 3 over 5 close parentheses to the i minus 1 power from 1 to infinity is a convergent or divergent geometric series and find the sum, if possible.identify whether the series summation of 12 open parentheses 3 over 5 close parentheses to the i minus 1 power from 1 to infinity is a convergent or divergent geometric series and find the sum, if possible.
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Ответ:
Given series is sigma i=1 to infinity 12*
,
=![12+ 12*\frac{3}{5} + 12*\frac{3}{5}^{2}+.......](/tpl/images/0073/8084/3e0fe.png)
Clearly it is a geometric series and it converges if and only if r<1
So, common ratio =
< 1
Hence the series is convergent.
Formula for sum of infinite geometric series =![\frac{a}{1-r}](/tpl/images/0073/8084/bb155.png)
Where a is the first term and r is the common ratio.
So, sum of series =![\frac{12}{1-\frac{3}{5} } = \frac{12}{\frac{2}{5} }](/tpl/images/0073/8084/86121.png)
For taking 2/5 to numerator we have to multiply with reciprocal 5/2 on both numerator and denominator.
Hence sum of series = 12*
= 30
Ответ: