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Math 121. Sequences and Summation Notation (Section 11.1) Names. 1. Find the first three terms and 6th term of a sequence whose nth term is given by an = (−1)n−1 1 − 9n Solution: The first three terms are a1 = (−1)1−1 1 1 = =− 1 − 9(1) −8 8 a2 = (−1)2−1 −1 1 = = 1 − 9(2) −17 17 a3 = (−1)3−1 1 1 = =− 1 − 9(3) −26 26 and the 6th term is a6 = −1 1 (−1)6−1 = = 1 − 9(6) −53 53 2. Find the 34th term of the sequence whose nth term is defined by an = (n + 2)! , n≥1 (n − 1)! Solution: The 34th term is (notice there is a lot of cancellation!) a34 = 36! 36 · 35 · 34 · 33 · 32 · · · 3 · 2 · 1 (34 + 2)! = = = 36 · 35 · 34 = 42840 (34 − 1)! 33! 33 · 32 · · · 3 · 2 · 1 3. Evaluate the following sum 7 X (−1)n (5n) n=4 Solution: The expanded sum is (−1)4 (5)(4) + (−1)5 (5)(5) + (−1)6 (5)(6) + (−1)7 (5)(7) which simplifies to −10. Fall 2016 4. (a) Write an expression for an , the nth term of the sequence of even numbers (b) Write an expression for bn , the nth term of the sequence of odd numbers 2, 4, 6, 8, 10, 12 . . . 1, 3, 5, 7, 9, 11, . . . (c) Write an expression for cn , the nth term of the sequence whose first six terms are 5, 7, 9, 11, 13, 15, . . . (d) Write an expression for dn , the nth term of the sequence whose first six terms are 6, − 8, 10, − 12, 14, − 16, . . . (e) Write an expression for en , the nth term of the sequence whose terms are multiples of 7, so its first six terms are 7, 14, 21, 28, 35, 42, . . . (f) Write an expression for fn , the nth term of the sequence whose first six terms are 4, 11, 18, 25, 32, 39, . . . Solution: (a) an = 2n (b) bn = 2n − 1 (c) Each cn results from adding 3 to each even number, so cn = 2n + 3. (d) The absolute value of each dn is obtained by adding 4 to each even number, and the signs of dn are negative on the even terms, so dn = (−1)n+1 (2n + 4). (e) en = 7n (f) The terms fn are obtained by subtracting 3 from each en and so fn = 7n − 3. 5. Express the following sum in summation notation. 5 6 7 8 + + + 9 16 25 36 Solution: Observe the numerator is n + 4 while the denominator is (n + 2)2 for n = 1, . . . , 4. Therefore, 4 X 6 7 8 n+4 5 + + + = 9 16 25 36 n=1 (n + 2)2 There are other ways to express this sum. Page 2