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How do you find the sum of #2/1+4/3+8/9+16/27+...+2^(n+1)/3^n+...#?

Answer 1

To find the sum of ( \frac{2}{1} + \frac{4}{3} + \frac{8}{9} + \frac{16}{27} + \ldots + \frac{2^{n+1}}{3^n} + \ldots ), we can recognize that each term is a geometric series. The common ratio between consecutive terms is ( \frac{2}{3} ). Using the formula for the sum of an infinite geometric series, which is ( \frac{a}{1 - r} ), where ( a ) is the first term and ( r ) is the common ratio, we can calculate the sum. So, the sum of the given series is ( \frac{\frac{2}{1}}{1 - \frac{2}{3}} ). Simplifying this expression yields ( \frac{6}{1 - \frac{2}{3}} ), which further simplifies to ( \frac{6}{\frac{1}{3}} ). Thus, the sum of the series is ( 18 ).

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Answer 2

Samantha Adkison

#sum_(n=0)^oo 2^(n+1)/3^n = 6#

#sum_(n=0)^oo 2^(n+1)/3^n = sum_(n=0)^oo (2 xx 2^n)/3^n#

#sum_(n=0)^oo 2^(n+1)/3^n = 2 sum_(n=0)^oo ( 2/3)^n#

This is the sum of a geometric series of ratio: #q=2/3#, and we know that for #q < 1#

#sum_(n=0)^oo q^n = 1/(1-q)#

so:

#sum_(n=0)^oo 2^(n+1)/3^n = 2 (1/(1-2/3)) = 6#

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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.