Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

Section 7.1 Answers

  • Page ID
    29447
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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. \(R = 1\\: I = (0, 2)\) 

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