How do you find the sum of the infinite geometric series 12+4+4/3+...?

Answer 1

#18#

An infinite geometric series of the form #sum_(n=1)^ooar^(n-1) # converges if and only if #|r|<1#, where r is the common ration between terms and given by #r=x_(n+1)/(x_n)#. In this case it converges to the value #a/(1-r)# where a is the first term in the corresponding sequence #(x_n)#.
So in this case, #r=4/12=4/3/4=1/3<1# hence the series converges.
Its sum is hence #sum_(n=1)^oo12*(1/3)^(n-1)=12/(1-1/3)=18#
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Answer 2

To find the sum of an infinite geometric series, use the formula for the sum:

[ S = \frac{a}{1 - r} ]

Where:

  • ( a ) is the first term of the series
  • ( r ) is the common ratio

For the given series ( 12 + 4 + \frac{4}{3} + \ldots ), the first term ( a = 12 ) and the common ratio ( r = \frac{1}{3} ).

Plug these values into the formula:

[ S = \frac{12}{1 - \frac{1}{3}} ]

[ S = \frac{12}{\frac{2}{3}} ]

[ S = 18 ]

So, the sum of the infinite geometric series ( 12 + 4 + \frac{4}{3} + \ldots ) is ( 18 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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