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Chapter 13: Sequences & Series

Chapter 13: Sequences & Series. L13.3 Arithmetic & Geometric Series and Their Sums. Sequences and Series. A sequence is an ordered progression of terms: t 1 , t 2 , t 3 , t 4 , …, t n A series is the sum of a sequence, i.e., a number:

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Chapter 13: Sequences & Series

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  1. Chapter 13: Sequences & Series L13.3 Arithmetic & Geometric Series and Their Sums

  2. Sequences and Series • A sequence is an ordered progression of terms: t1, t2, t3, t4, …, tn • A series is the sum of a sequence, i.e., a number: t1 + t2 + t3 + t4 + …+ tn • Here we study finite series, i.e., n is a real number. • Sn is defined to be the sum of a series with n terms Defined explicitly, Sn = t1 + t2 + t3 + t4 + …+ tn Defined recursively, S0 = 0, Sn = Sn-1 + tn, n > 0

  3. Finite Arithmetic & Geometric Serieshave formulas to compute sums * • Sum of a finite Arithmetic Series: • Sum of a finite Geometric Series: • Examples: • Find the sum of the first 25 terms of the series 11 + 14 + 17 + 20 + … • Find the sum of the first 10 terms of the series 1 + 1/3 + 1/9 + 1/27 + … • Find the sum of the multiples of 7 between 100 and 1000 • Find the sum of 14 + 24 + 34 + 44 + … + 104 Arithmetic, d = 3; t25 = 11 + (25 – 1)3 = 83, So S25 = 25(11 + 83)/2 = 1175 Geometric, r = ⅓, So S10 = 1(1 – (⅓)10)/(1 – ⅓) ≈ 1.5 Arithmetic, t1 = 105, tn = 994, n = 128, So S128 = 128(105 + 994)/2 = 70,336 Neither, must be done recursively. …… S9 = S8 + 94,S10 = S9 + 104 = 25,333 * You are responsible for the proofs of these two formulas. See pages 487 & 488.

  4. Finite Arithmetic & Geometric Series • Sums are just numbers so can be added, subtracted, etc. • Final example: Find the sum of 1 − 4 + 7 − 10 + 13 − 16 + 19 − …+ 1003 This series is neither arithmetic nor geometric but can be decomposed into 2 arithmetic ones: Orig: 1 − 4 + 7 − 10 + 13 − 16 + 19 − … + 1003 1 + 7 + 13 + 19 + … + 1003 − (4 + 10 + 16 + … + 1000) The first series has t1=1, tn = 1003, d = 6. Need n. 1003 = 1 + (n – 1)6; n = 168 The second series as t1=4, tn = 1000, d = 6, Need n. 1000 = 4 + (n – 1)6; n = 167 Therefore, S(168+167) = S168 – S167 =

  5. Finite Arithmetic & Geometric SeriesFormula Derivations • Sum of a finite Arithmetic Series: Write Sn twice, the second time with the order reversed: Sn = t1 + (t1 + d) + (t1 + 2d) + (t1 + 3d) + … + (tn – d) + tn Sn = tn + (tn – d) + (tn – 2d) + (tn – 3d) + … + (t1 + d) + t1 2Sn = (t1 + tn) + (t1 + tn) + (t1 + tn) + (t1 + tn) + … + (t1 + tn) + (t1 + tn) 2Sn = n(t1 + tn) • Sum of a finite Geometric Series: Multiply Sn by the common ratio and subtract the result from Sn: Sn = t1 + t1r + t1r2 + t1r3 + … + t1rn-1 rSn = t1r + t1r2 + t1r3 + … + t1rn-1 + t1rn Sn – rSn = t1 – t1rn = t1(1 – r)

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