A.7.1: Section 7.1 Answers
- Page ID
- 43776
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- \(R = 2;\: I = (−1, 3)\)
- \(R = 1/2;\: I = (3/2, 5/2)\)
- \(R = 0\)
- \(R = 16;\: I = (−14, 18)\)
- \(R = ∞;\: I = (−∞, ∞)\)
- \(R = 4/3;\: I = (−25/3, −17/3)\)
3.
- \(R = 1;\: I = (0, 2)\)
- \(R = \sqrt{2};\: I = (−2 −\sqrt{2}, −2 + \sqrt{2})\)
- \(R = ∞;\: I = (−∞,∞)\)
- \(R = 0\)
- \(R = \sqrt{3};\: I = (− \sqrt{3}, \sqrt{3})\)
- \(R = 1\\[4pt]: I = (0, 2)\)
5.
- \(R = 3;\: I = (0, 6)\)
- \(R = 1;\: I = (−1, 1)\)
- \(R = 1/\sqrt{3};\: I = (3 − 1/\sqrt{3}, 3 + 1/\sqrt{3})\)
- \(R = ∞;\: I = (−∞, ∞)\)
- \(R = 0\)
- \(R = 2;\: I = (−1, 3)\)
11. \(b_{n} = 2(n + 2)(n + 1)a_{n+2} + (n + 1)na_{n+1} + (n + 3)a_{n}\)
12. \(b_{0} = 2a_{2} − 2a_{0}\: b_{n} = (n + 2)(n + 1)a_{n+2} + [3n(n − 1) − 2]a_{n} + 3(n − 1)a_{n−1},\: n ≥ 1\)
13. \(b_{n} = (n + 2)(n + 1)a_{n+2} + 2(n + 1)a_{n+1} + (2n^{2} − 5n + 4)a_{n}\)
14. \(b_{n} = (n + 2)(n + 1)a_{n+2} + 2(n + 1)a_{n+1} + (n^{2} − 2n + 3)a_{n}\)
15. \(b_{n} = (n + 2)(n + 1)a_{n+2} + (3n^{2} − 5n + 4)a_{n}\)
16. \(b_{0} = −2a_{2} + 2a_{1} + a_{0},\: b_{n} = −(n + 2)(n + 1)a_{n+2} + (n + 1)(n + 2)a_{n+1} + (2n + 1)a_{n} + a_{n−1},\: n ≥ 2\)
17. \(b_{0} = 8a_{2} + 4a_{1} − 6a_{0},\: b_{n} = 4(n + 2)(n + 1)a_{n+2} + 4(n + 1)^{2}a_{n+1} + (n^{2} + n − 6)a_{n} − 3a_{n−1},\: n ≥ 1\)
21. \(b_{0} = (r + 1)(r + 2)a_{0},\: b_{n} = (n + r + 1)(n + r + 2)a_{n} − (n + r − 2)^{2} a_{n−1},\: n ≥ 1.\)
22. \(b_{0} = (r − 2)(r + 2)a_{0},\: b_{n} = (n + r − 2)(n + r + 2)a_{n} + (n + r + 2)(n + r − 3)a_{n−1},\: n ≥ 14\)
23. \(b_{0} = (r − 1)^{2} a_{0},\: b_{1} = r^{2}a_{1} + (r + 2)(r + 3)a_{0},\: bn = (n + r − 1)^{2} a_{n} + (n + r + 1)(n + r + 2)a_{n−1} + (n + r − 1)a_{n−2},\: n ≥ 2\)
24. \(b_{0} = r(r + 1)a_{0},\: b_{1} = (r + 1)(r + 2)a_{1} + 3(r + 1)(r + 2)a_{0},\: b_{n} = (n + r)(n + r + 1)a_{n} + 3(n + r)(n + r + 1)a_{n−1} + (n + r)a_{n−2},\: n ≥ 2\)
25. \(b_{0} = (r + 2)(r + 1)a_{0},\: b_{1} = (r + 3)(r + 2)a_{1},\: b_{n} = (n + r + 2)(n + r + 1)a_{n} + 2(n + r − 1)(n + r − 3)a_{n−2},\: n ≥ 2\)
26. \(b_{0} = 2(r + 1)(r + 3)a_{0},\: b_{1} = 2(r + 2)(r + 4)a_{1},\: b_{n} = 2(n + r + 1)(n + r + 3)a_{n} + (n + r − 3)(n + r)a_{n−2},\: n ≥ 2\)