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Find the sum. (a) 1/3+ 2/3² + {2²}/{3³}+{2³}/{3⁴}+ ...... + {2 ⁹}/{3¹⁰}

User Kett
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Final answer:

To find the sum of the series 1/3+ 2/3² + {2²}/{3³}+{2³}/{3⁴}+ ...... + {2⁹}/{3¹⁰}, use the formula for the sum of a geometric series with the first term 1/3 and common ratio 2/3, assuming there are 10 terms.

Step-by-step explanation:

The sum of the series given is 1/3+ 2/3² + {2²}/{3³}+{2³}/{3⁴}+ ...... + {2⁹}/{3¹⁰}. This is a geometric series where the first term (a) is 1/3 and the common ratio (r) is 2/3. We can use the formula for the sum of the first n terms of a geometric series, which is a(1-r^n)/(1-r) to find the sum of this series.

The series in question does not have a finite number of terms specified, but it seems to end at the term {2⁹}/{3¹⁰}, which would make it the 10th term. Assuming this interpretation, we can calculate the sum:

  1. First term (a) = 1/3
  2. Common ratio (r) = 2/3
  3. Number of terms (n) = 10
  4. Sum (S) = a(1-r^n)/(1-r) = (1/3)(1-(2/3)^10)/(1-(2/3))

Simplifying this, we get the sum of the series. However, the actual process of simplifying would involve calculating (2/3)^10 and then performing the subtraction and division as indicated in the formula.

User Bennyxguo
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