Concepts Examples nth Term an =a1 + 1n −12d Sum of the First n Terms Sn = n 2 1 a1 +an2 or Sn = n 2 3 2a1 + 1n −12d4 For the arithmetic sequence 2, 5, 8, 11, c , suppose that n = 10. The 10th term is found as follows. a10 = 2 + 110 - 123 a1 = 2, n = 10, d = 3 a10 = 2 + 9 # 3 a10 = 29 The sum of the first 10 terms can be found in either of two ways. S10 = 10 2 1a1 + a102 S10 = 512 + 292 S10 = 155 or S10 = 10 2 32122 + 110 - 1234 S10 = 514 + 9 # 32 S10 = 514 + 272 S10 = 155 Evaluate 4!. 4! = 4 # 3 # 2 # 1 = 24 For the geometric sequence 1, 2, 4, 8, c , a1 = 1. r = 8 4 = 2 Common ratio (Any two successive terms could have been used.) Suppose that n = 6. Then the sixth term is a6 = 1121226-1 = 11225 = 32. The sum of the first six terms is found as follows. S6 = 111 - 262 1 - 2 = 1 - 64 -1 = 63 The sum of the terms of the infinite geometric sequence a∞ k=0 a 1 2b k = 1 + 1 2 + 1 4 + g is found as follows. S∞ = 1 1 - 1 2 = 1 1 2 = 2 11.3 Geometric Sequences and Series Assume a1 is the first term, an is the nth term, and r is the common ratio in a geometric sequence. Common Ratio r = an+1 an nth Term an =a1r n−1 Sum of the First n Terms Sn = a111 −r n2 1 −r , where r 31 Sum of the Terms of an Infinite Geometric Sequence SH = a1 1 −r , where 0 r 0 *1 11.4 The BinomialTheorem n-Factorial For any positive integer n, n! =n1n −12 1n −22 P132 122 112. By definition, 0! =1. 1113 CHAPTER 11 Test Prep
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