(a) \(r = \dfrac{{{u_2}}}{{{u_1}}} = \dfrac{{\;\;\dfrac{{x{{\left( {x + 1} \right)}^2}}}{{{{\left( {x - 2} \right)}^2}}}\;\;}}{{\dfrac{{x\left( {x + 1} \right)}}{{x - 2}}}} = \dfrac{{x{{\left( {x + 1} \right)}^2}}}{{{{\left( {x - 2} \right)}^2}}} \cdot \dfrac{{x - 2}}{{x\left( {x + 1} \right)}} = \dfrac{{x + 1}}{{x - 2}}\)(b) \({u_n} = {u_1}{r^{n - 1}} = \dfrac{{x\left( {x + 1} \right)}}{{x - 2}}{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)^{n - 1}} = \dfrac{{x{{\left( {x + 1} \right)}^1}{{\left( {x + 1} \right)}^{n - 1}}}}{{{{\left( {x - 2} \right)}^1}{{\left( {x - 2} \right)}^{n - 1}}}} = \dfrac{{x{{\left( {x + 1} \right)}^n}}}{{{{\left( {x - 2} \right)}^n}}}\) or \({u_n} = x{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)^n}\)(c) \({S_n} = \dfrac{{{u_1}\left( {{r^n} - 1} \right)}}{{r - 1}}\)\({S_n} = \dfrac{{x\left( {x + 1} \right)\left[ {{{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)}^n} - 1} \right]}}{{\left( {x - 2} \right)\left( {\dfrac{{x + 1}}{{x - 2}} - 1} \right)}} = \dfrac{{x\left( {x + 1} \right)\left[ {{{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)}^n} - 1} \right]}}{{\left( {x - 2} \right)\left( {\dfrac{{x + 1}}{{x - 2}} - \dfrac{{x - 2}}{{x - 2}}} \right)}}\)\( = \dfrac{{x\left( {x + 1} \right)\left[ {{{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)}^n} - 1} \right]}}{{\left( {x - 2} \right)\left( {\dfrac{3}{{x - 2}}} \right)}} = \dfrac{{x\left( {x + 1} \right)\left[ {{{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)}^n} - 1} \right]}}{3}\)Thus, \({S_n} = \dfrac{x}{3}\left( {x + 1} \right)\left[ {\dfrac{{{{\left( {x + 1} \right)}^n}}}{{{{\left( {x - 2} \right)}^n}}} - 1} \right]\) Q.E.D.(d) For an infinite sum to exist, it must be that \(\left| r \right| < 1\) which is equivalent to \( - 1 < r < 1\) or \({r^2} < 1\).\({r^2} < 1\;\;\;\; \Rightarrow \;\;\;\;{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)^2} < 1\;\;\;\; \Rightarrow \;\;\;\;\dfrac{{{x^2} + 2x + 1}}{{{x^2} - 4x + 2}} < 1\)Since \({x^2} - 4x + 2 \ge 0\) for all values of x, then we can multiply both sides of the inequality and the inequality sign is not reversed.\(\dfrac{{{x^2} + 2x + 1}}{{{x^2} - 4x + 2}} < 1\;\;\;\; \Rightarrow \;\;\;\;{x^2} + 2x + 1 < {x^2} - 4x + 2\;\;\;\; \Rightarrow \;\;\;\;6x < 3\;\;\;\; \Rightarrow \;\;\;\;x < \dfrac{1}{2}\)Thus, the sequence has an infinite sum when \(x < \dfrac{1}{2}\).