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Geometric Series (page 5 of 5) Sections: Terminology and notation, Basic examples, Arithmetic and geometric sequences, Arithmetic series, Finite and infinite geometric series You can take the sum of a finite number of terms of a geometric sequence. And, for reasons you'll study in calculus, you can take the sum of an infinite geometric sequence, but only in the special circumstance that the common ratio r is between –1 and 1; that is, you have to have  r  < 1. For a geometric sequence with first term a_{1} = a and common ratio r, the sum of the first n terms is given by: Note: Your book may have a slightly different form of the partialsum formula above. For instance, the "a" may be multiplied through the numerator, the factors in the fraction might be reversed, or the summation may start at i = 0 and have a power of n + 1 on the numerator. All of these forms are equivalent, and the formulation above may be derived from polynomial long division. In the special case that  r  < 1, the infinite sum exists and has the following value: Copyright © Elizabeth Stapel 20062011 All Rights Reserved
The first few terms are –6, 12, –24, so this is a geometric series with common ratio r = –2. (I can also tell that this must be a geometric series because of the form given for each term: as the index increases, each term will be multiplied by an additional factor of –2.) The first term of the sequence is a = –6. Plugging into the summation formula, I get: So the value of the summation is 2 097 150 The notation "S_{10}" means that I need to find the sum of the first ten terms. The first term is a = 250. Dividing pairs of terms, I get 100 ÷ 250 = 2/5, 40 ÷ 100 = 2/5, etc, so the terms being added form a geometric sequence with common ratio r = 2/5. When I plug in the values of the first term and the common ratio, the summation formula gives me:
They've given me the sum of the first four terms, S_{4}, and the value of the common ratio r. Since there is a common ratio, I know this must be a geometric series. Plugging into the geometricseriessum formula, I get: Multiplying on both sides by 27/40 to solve for the first term a = a_{1}, I get: Then:
There's a trick to this. I first have to break the repeating decimal into separate terms: 0.333... = 0.3 + 0.03 + 0.003 + 0.0003 + ... This shows the repeating pattern of the nonterminating (neverending) decimal explicitly: For each term, I have a decimal point, followed by a steadilyincreasing number of zeroes, and then ending with a "3". This can be written in fractional form, and then converted into geometricseries form: Then 0.333... is an infinite geometric series with a = 3/10 and r = 1/10. Since  r  < 1, I can use the formula for summing infinite geometric series: Using the summation formula to show that the geometric series "expansion" of 0.333... has a value of onethird is the "showing" that the exercise asked for. You can use this method to convert any repeating decimal to its fractional form:
First I'll break this into its constituent parts, so I can find the pattern: 1.363636.. = 1 + 0.36 + 0.0036 + 0.000036 + ... There are two digits that repeat, so the fractions are a little bit different. But this is still a geometric series: Then this is the leading "1", plus a geometric series having a = 9/25 and r = 1/100. Then the sum is: Note: This technique can also be used to convert any repeating decimal into fractional form, and also can be used to prove that 0.999... = 1. << Previous Top  1  2  3  4  5  Return to Index



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