A geometric seriesThe sum of the terms of a geometric sequence. is the sum of the terms of a geometric sequence. For example, the sum of the first 5 terms of the geometric sequence defined by an=3n+1 follows: Show S5=Σn=153n+1=31+1+32+1+33+1+34+1+35+1=32+33+34+35+36=9+27+81+243+729=1,089 Adding 5 positive integers is managable. However, the task of adding a large number of terms is not. Therefore, we next develop a formula that can be used to calculate the sum of the first n terms of any geometric sequence. In general, Sn=a1+a1r+a1r2+…+a1rn−1 Multiplying both sides by r we can write, rSn= a1r+a1r2+a1r3+…+a1rn Subtracting these two equations we then obtain, Sn−rSn=a1−a1rnSn(1−r)=a1(1−rn) Assuming r≠1 dividing both sides by (1−r) leads us to the formula for the nth partial sum of a geometric sequenceThe sum of the first n terms of a geometric sequence, given by the formula: Sn=a1(1−rn)1−r, r≠1.: Sn=a1(1−rn)1−r(r≠1) In other words, the nth partial sum of any geometric sequence can be calculated using the first term and the common ratio. For example, to calculate the sum of the first 15 terms of the geometric sequence defined by an=3n+1, use the formula with a1=9 and r=3. S15=a1(1−r15)1−r=9⋅(1−315)1−3=9(−14,348,906)−2=64,570,077 Example 4Find the sum of the first 10 terms of the given sequence: 4, −8, 16, −32, 64,… Solution: Determine whether or not there is a common ratio between the given terms. r=−84=−2 Note that the ratio between any two successive terms is −2; hence, the given sequence is a geometric sequence. Use r=−2 and the fact that a1=4 to calculate the sum of the first 10 terms, Sn=a1(1−rn)1−rS10=4[1−(−2)10]1−(−2)=4(1−1,024)1+2=4(−1,023)3=−1,364 Answer: S10=−1,364 Example 5Evaluate: Σn=162(−5)n. Solution: In this case, we are asked to find the sum of the first 6 terms of a geometric sequence with general term an=2(−5)n. Use this to determine the 1st term and the common ratio r: a1=2(−5)1=−10 To show that there is a common ratio we can use successive terms in general as follows: r=anan−1=2(−5)n2(−5)n−1=(−5)n−(n−1)=(−5)1=−5 Use a1=−10 and r=−5 to calculate the 6th partial sum. Sn=a1(1−rn)1−rS6=−10[1−(−5)6]1−(−5)=−10(1−15,625)1+5=−10(−15,624)6=26,040 Answer: 26,040 Try this! Find the sum of the first 9 terms of the given sequence: −2, 1, −1/2,… Answer: S9=−171128 If the common ratio r of an infinite geometric sequence is a fraction where |r|<1 (that is −1<r<1), then the factor (1−rn) found in the formula for the nth partial sum tends toward 1 as n increases. For example, if r=110 and n=2,4,6 we have, 1−(110)2=1−0.01=0.991−(110)4=1−0.0001=0.99991−(110)6=1−0.000001=0.999999 Here we can see that this factor gets closer and closer to 1 for increasingly larger values of n. This illustrates the idea of a limit, an important concept used extensively in higher-level mathematics, which is expressed using the following notation: limn→∞(1−rn)=1 where |r|<1 This is read, “the limit of (1−rn) as n approaches infinity equals 1.” While this gives a preview of what is to come in your continuing study of mathematics, at this point we are concerned with developing a formula for special infinite geometric series. Consider the nth partial sum of any geometric sequence, Sn=a1(1−rn)1−r=a11−r(1−rn) If |r|<1 then the limit of the partial sums as n approaches infinity exists and we can write, Sn=a11−r(1−rn) ⇒n→∞ S∞=a11−r⋅1 Therefore, a convergent geometric seriesAn infinite geometric series where |r|<1 whose sum is given by the formula: S∞=a11−r. is an infinite geometric series where |r|<1; its sum can be calculated using the formula: S∞=a11−r Example 6Find the sum of the infinite geometric series: 32+12+16+118+154+⋯ Solution: Determine the common ratio, r=1232=12⋅23=13 Since the common ratio r=13 is a fraction between −1 and 1, this is a convergent geometric series. Use the first term a1=32 and the common ratio to calculate its sum. S∞=a11−r=321−(13)= 32 23=32⋅32=94 Answer: S∞=94 Note: In the case of an infinite geometric series where |r|≥1, the series diverges and we say that there is no sum. For example, if an=(5)n−1 then r=5 and we have S∞=Σn=1∞(5)n−1=1+5+25+⋯ We can see that this sum grows without bound and has no sum. Try this! Find the sum of the infinite geometric series: Σn=1∞−2(59)n−1. Answer: −9/2 A repeating decimal can be written as an infinite geometric series whose common ratio is a power of 1/10. Therefore, the formula for a convergent geometric series can be used to convert a repeating decimal into a fraction. Example 7Write as a fraction: 1.181818… Solution: Begin by identifying the repeating digits to the right of the decimal and rewrite it as a geometric progression. 0.181818…=0.18+0.0018+0.000018+…=18100+1810,000+181,000,000+… In this form we can determine the common ratio, r=1810,00018100=1810,000×10018=1100 Note that the ratio between any two successive terms is 1100. Use this and the fact that a1=18100 to calculate the infinite sum: S∞=a11−r=181001−(1100)=1810099100=18100⋅10099=211 Therefore, 0.181818…=211 and we have, 1.181818…=1+211=1211 Answer: 1211 Example 8A certain ball bounces back to two-thirds of the height it fell from. If this ball is initially dropped from 27 feet, approximate the total distance the ball travels. Solution: We can calculate the height of each successive bounce: 27⋅23=18 feet Height of the first bounce18⋅23=12 feet Height of the second bounce12⋅23=8 feet Height of the third bounce The total distance that the ball travels is the sum of the distances the ball is falling and the distances the ball is rising. The distances the ball falls forms a geometric series, 27+18+12+⋯ Distance the ball is falling where a1=27 and r=23. Because r is a fraction between −1 and 1, this sum can be calculated as follows: S∞=a11−r=271−23=2713=81 Therefore, the ball is falling a total distance of 81 feet. The distances the ball rises forms a geometric series, 18+12+8+⋯ Distance the ball is rising where a1=18 and r=23. Calculate this sum in a similar manner: S∞=a11−r=181−23=1813=54 Therefore, the ball is rising a total distance of 54 feet. Approximate the total distance traveled by adding the total rising and falling distances: 81+54=135 feet Answer: 135 feet Key Takeaways
Topic Exercises
Part A: Geometric SequencesWrite the first 5 terms of the geometric sequence given its first term and common ratio. Find a formula for its general term. Given the geometric sequence, find a formula for the general term and use it to determine the 5th term in the sequence. Given the terms of a geometric sequence, find a formula for the general term. Find all geometric means between the given terms.
Part B: Geometric SeriesCalculate the indicated sum. Write as a mixed number. Part C: Sequences and SeriesCategorize the sequence as arithmetic, geometric, or neither. Give the common difference or ratio, if it exists. Categorize the sequence as arithmetic or geometric, and then calculate the indicated sum. Calculate the indicated sum.
Part D: Discussion Board |