# 11.43: A.7.1- Section 7.1 Answers

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1.

1. $$R = 2;\: I = (−1, 3)$$
2. $$R = 1/2;\: I = (3/2, 5/2)$$
3. $$R = 0$$
4. $$R = 16;\: I = (−14, 18)$$
5. $$R = ∞;\: I = (−∞, ∞)$$
6. $$R = 4/3;\: I = (−25/3, −17/3)$$

3.

1. $$R = 1;\: I = (0, 2)$$
2. $$R = \sqrt{2};\: I = (−2 −\sqrt{2}, −2 + \sqrt{2})$$
3. $$R = ∞;\: I = (−∞,∞)$$
4. $$R = 0$$
5. $$R = \sqrt{3};\: I = (− \sqrt{3}, \sqrt{3})$$
6. $$R = 1\\[4pt]: I = (0, 2)$$

5.

1. $$R = 3;\: I = (0, 6)$$
2. $$R = 1;\: I = (−1, 1)$$
3. $$R = 1/\sqrt{3};\: I = (3 − 1/\sqrt{3}, 3 + 1/\sqrt{3})$$
4. $$R = ∞;\: I = (−∞, ∞)$$
5. $$R = 0$$
6. $$R = 2;\: I = (−1, 3)$$

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. $$b_{0} = 2(r + 1)(r + 3)a_{0},\: b_{1} = 2(r + 2)(r + 4)a_{1},\: b_{n} = 2(n + r + 1)(n + r + 3)a_{n} + (n + r − 3)(n + r)a_{n−2},\: n ≥ 2$$

This page titled 11.43: A.7.1- Section 7.1 Answers is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.