### Geometric sequence; log; reduce to quadratic

Question Sample Titled 'Geometric sequence; log; reduce to quadratic'

A fund ${A}$ can make a revenue of ${A}{\left({n}\right)}$ dollars in the ${n}^{{{t}{h}}}$ year since the start of investment, where ${n}$ is a positive integer. It is given that ${A}{\left({n}\right)}={4}{p}{r}^{{n}}$ , where ${p}$ and ${r}$are positive constants. It is found that the revenue made by fund ${A}$ in the ${1}^{{{s}{t}}}$ and the ${2}^{{{n}{d}}}$ year since the start of investment are $3520000$ and $3872000$ respectively.

 (a) (i) Find ${p}$ and ${r}$ . (ii) Express the total revenue made by fund ${A}$ in the first ${n}$ years since the start investment in terms of ${n}$ .  (5 marks)  (b) Aaron plans to invest fund ${A}$ and he plans to start investing in a new fund ${B}$ after fund ${A}$ investment has been started for ${4}$ years. Let ${B}{\left({m}\right)}$ dollars be the revenue made by the fund ${B}$ in the ${m}^{{{t}{h}}}$ year since the start of its investment, where ${m}$ is a positive integer. It is given that ${B}{\left({m}\right)}={p}{r}^{{{2}{m}}}$ . (i) The fund manager claims that after Aaron has started investing fund ${B}$, the yearly revenue made by fund ${B}$ would exceed fund ${A}$ with ${17}$ years. Do you agree? Explain your answer. (ii) Aaron thinks that when the total revenue made by fund ${A}$ and fund ${B}$ since the start of the investment in fund ${A}$ exceeds $521000000$521000000$ , he should stop all the investments. In which year after the start of the investment in fund ${A}$ should Aaron stop all the investments?  (7 marks) 題解  (ai) ${4}{p}{r}$ $={3520000}$ $\ldots{\left({1}\right)}$ ${4}{p}{r}^{{2}}$ $={3872000}$ $\ldots{\left({2}\right)}$ ${\left({2}\right)}\div{\left({1}\right)}:$ ${r}$ $=\dfrac{{3872000}}{{3520000}}$  $={1.1}$ 1A $\therefore{p}$ $={800000}$ 1A  (aii) The total revenue made by ${A}$ in the first ${n}$ years  $={4}{p}{r}+{4}{p}{r}^{{2}}+{4}{p}{r}^{{3}}+\ldots+{4}{p}{r}^{{n}}$ 1M  $={4}{p}{r}\times\dfrac{{{r}^{{n}}-{1}}}{{{r}-{1}}}$ 1M  =$35200000(rn−1)$=35200000\left({r}^{n}-1\right)$ 1A  (bi) $\dfrac{{{B}{\left({m}\right)}}}{{{A}{\left({n}\right)}}}$ $=\dfrac{{{p}{r}^{{{2}{m}}}}}{{{4}{p}{r}^{{n}}}}$  $=\dfrac{{{r}^{{{2}{\left({n}-{4}\right)}}}}}{{{4}{r}^{{n}}}}$  $=\dfrac{{{r}^{{{n}-{8}}}}}{{4}}$  $=\dfrac{{{r}^{{{n}-{8}}}}}{{4}}$  Let $\dfrac{{{B}{\left({m}\right)}}}{{{A}{\left({n}\right)}}}$ $>{1}$ , 1M $\dfrac{{{r}^{{{n}-{8}}}}}{{4}}$ $>{1}$ ${r}^{{{n}-{8}}}$ $>{4}$ ${{\log{{r}}}^{{{n}-{8}}}}$ $>{\log{{4}}}$ ${\left({n}-{8}\right)}{\log{{1.1}}}$ $>{\log{{4}}}$ ${n}$ $>\dfrac{{{\log{{4}}}}}{{{\log{{1.1}}}}}+{8}$ ${n}$ $>{22.545081794683426}$ 1A $\therefore$The yearly revenue made by fund ${B}$ would exceed ${A}$ in the ${23}^{{{r}{d}}}$ year since the start of the investment in fund ${A}$ . 1A i.e. the ${19}^{{{t}{h}}}$ year since the start of the investment in fund ${B}$ . Thus, the manager is incorrect.  (bii) Let ${k}={35200000}$ , ${x}={r}^{{n}}$ . The total revenue made by ${A}$ and ${B}$ since the start of the investment in ${A}$ $={k}{\left({r}^{{n}}-{1}\right)}+{\left({p}{r}^{{2}}+{p}{r}^{{4}}+\ldots+{p}{r}^{{{2}{\left({n}-{4}\right)}}}\right)}$ $={k}{\left({r}^{{n}}-{1}\right)}+\dfrac{{{p}{r}^{{2}}{\left({r}^{{{2}{\left({n}-{4}\right)}}}-{1}\right)}}}{{{r}^{{2}}-{1}}}$ 1M $={k}{r}^{{n}}-{k}+\dfrac{{{p}{r}^{{{2}{n}-{6}}}}}{{{r}^{{2}}-{1}}}-\dfrac{{1}}{{{r}^{{2}}-{1}}}$ $={k}{r}^{{n}}-{k}+\dfrac{{{p}{r}^{{{2}{n}}}}}{{{r}^{{6}}{\left({r}^{{2}}-{1}\right)}}}-\dfrac{{1}}{{{r}^{{2}}-{1}}}$ $={k}{x}-{k}+\dfrac{{{p}{x}^{{2}}}}{{{r}^{{6}}{\left({r}^{{2}}-{1}\right)}}}-\dfrac{{1}}{{{r}^{{2}}-{1}}}$ $={\left(\dfrac{{p}}{{{r}^{{6}}{\left({r}^{{2}}-{1}\right)}}}\right)}{x}^{{2}}+{k}{x}-{k}-\dfrac{{1}}{{{r}^{{2}}-{1}}}$  Solve ${\left(\dfrac{{p}}{{{r}^{{6}}{\left({r}^{{2}}-{1}\right)}}}\right)}{x}^{{2}}+{k}{x}-{k}-\dfrac{{1}}{{{r}^{{2}}-{1}}}>{521000000}$ 1M ${\left(\dfrac{{p}}{{{r}^{{6}}{\left({r}^{{2}}-{1}\right)}}}\right)}{x}^{{2}}+{k}{x}-{k}-\dfrac{{1}}{{{r}^{{2}}-{1}}}-{521000000}>{0}$ By substitution and the quadratic formula, we have ${x}>{9.860891990119}$ or ${x}<-{26.230115630119}$ (rejected) 1A Thus, ${1.1}^{{n}}>{9.860891990119}$ ${n}{\log{{1.1}}}>{\log{{9.860891990119}}}$ ${n}>{24.01188031279158}$ Aaron should stop all the investments in the ${25}^{{{t}{h}}}$ year after the start of the investment in fund ${A}$ . 1A

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