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11 SL notes 2015-16 Unit 1: Algebra Lesson 6: Sum of an infinite series The sum to infinity of a series is written 𝑆∞ or simply S. When r lies in the range −1 < 𝑟 < 1, larger and larger values of N give smaller and smaller values of 𝑟 𝑛 , so we say lim 𝑟 𝑛 = 0. 𝑛→∞ Example: The terms of the sequence generated by the rule 𝑢𝑛 = 5(0.1)𝑛−1 are: 5, 0.5, 0.05, 0.005, 0.0005,…. The terms are becoming very small. Hence, the value of (1 − 𝑟 𝑛 ) tends to 1, and so 𝑢1 (1 − 𝑟 𝑛 ) 𝑆𝑛 = (1 − 𝑟) 𝑢 1 becomes 𝑆𝑛 = (1−𝑟) for −1 < 𝑟 < 1. If |𝑟 > 1| then the terms keep getting bigger, so the series is divergent and the sum is ±∞. 5 5 5 Example: Find the sum of the infinite series 5 + 4 + 16 + 64 + ⋯ 1 𝑢1 = 5 and 𝑟 + 4. 𝑢 1 𝑆∞ = (1−𝑟) 5 𝑆∞ = 1 (1 − 4) 20 = 3 20 The sequence is said converge to 3 . This is its limiting sum. Questions: 1 1 1 1) Find the sum to infinity of the series 1 − 5 + 25 − 125 + ⋯ 1 Solution: 𝑢1 = 1, and 𝑟 = − 5 𝑆∞ = 𝑢1 1 5 = = (1 − 𝑟) (1 − (− 1)) 6 5 Express 0.12̇ as a rational number. Do this using the sum of an infinite series, not by other methods. 1 2 2 2 Solution: 0.12̇ = 10 + 100 + 1000 + 10000 + ⋯ 2) = 1 2 2 1 2 1 2 + + ( )+ ( ) +⋯ 10 100 100 10 100 10 1 2 1 Which is 10 plus a geometric series with 𝑢1 = 100 and 𝑟 = 10. 1 Hence 0.12̇ = 10 + 2 100 1 (1− ) 10 1 1 11 = 10 + 45 = 90, which you can check with a calculator. 11 SL notes 2015-16 3) The sum of an infinite G.P is three times its first term. Find its common ratio. 𝑢1 Solution: = 3𝑢1 (1−𝑟) ⟹ 𝑢1 = 3𝑢1 (1 − 𝑟) ⟹ 𝑢1 = 3𝑢1 − 3𝑢1 𝑟 ⟹ −3𝑢1 𝑟 = −2𝑢1 2 ⟹𝑟=3 4) A ball is dropped from a height of 1 metre and it always bounces to one half of its previous height. What is the total distance travelled by the ball? Solution: This is tricky because you need to remember that the ball drops one metre, then it bounces 50cm up and drops 50 cm down, then bounces 25 cm up and drops 25cm down……. 50 Distance travelled =100 + 2 ((1−0.5)) = 100 + 2(100) = 300𝑐𝑚 𝑜𝑟 3 𝑚𝑒𝑡𝑟𝑒𝑠. Page 170 ex 6G.2 Test on Sequences on Friday 18th September.