Write down the first three terms of the infinite series for \(G(t)\), the probability generating function for \(X\).

Determine the radius of convergence of this infinite series.

By considering \(\left( {1 - \frac{t}{3}} \right)G(t)\), show that

\[G(t) = \frac{{3kt}}{{{{(3 - t)}^2}}}.\]

Hence show that \(k = \frac{4}{3}\).

Show that \(\ln G(t) = \ln 4 + \ln t - 2\ln (3 - t)\).

By differentiating both sides of this equation, determine the values of \(G’(1)\) and \(G’’(1)\).

Hence find \({\text{Var}}(X)\).

\(G(t) = \frac{{kt}}{3} + \frac{{2k{t^2}}}{{{3^2}}} + \frac{{3k{t^3}}}{{{3^3}}} +  \ldots \)     M1A1

[??? marks]

\(\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{(n + 1)k{t^{n + 1}}}}{{{3^{n + 1}}}} \times \frac{{{3^n}}}{{nk{t^n}}}\)     M1A1

\( \to \frac{t}{3}{\text{ as }}n \to \infty \)     A1

for convergence, \(\left| {\frac{t}{3}} \right| < 1\) so radius of convergence \( = 3\)     A1

[??? marks]

\(G(t) = \frac{{kt}}{3} + \frac{{2k{t^2}}}{{{3^2}}} + \frac{{3k{t^3}}}{{{3^3}}} +  \ldots \)

\(\frac{t}{3}G(t) = \frac{{k{t^2}}}{{{3^2}}} + \frac{{2k{t^3}}}{{{3^3}}} +  \ldots \)

\(\left( {1 - \frac{t}{3}} \right)G(t) = \frac{{kt}}{3} + \frac{{k{t^2}}}{{{3^2}}} + \frac{{k{t^3}}}{{{3^3}}} +  \ldots \)     M1A1

\( = \frac{{\frac{{kt}}{3}}}{{\left( {1 - \frac{t}{3}} \right)}}\)     A1

\(G(t) = \frac{{\frac{{kt}}{3}}}{{{{\left( {1 - \frac{t}{3}} \right)}^2}}} = \frac{{3kt}}{{{{(3 - t)}^2}}}\)     AG

[??? marks]

\(G(1) = 1\)     M1

so \(k = \frac{4}{3}\)     AG

[??? marks]

\(\ln G(t) = \ln 4t - \ln {(3 - t)^2}\)     M1

\(\ln G(t) = \ln 4 + \ln t - 2\ln (3 - t)\)     AG

[??? marks]

\(\frac{{G'(1)}}{{G(1)}} = \frac{1}{t} + \frac{2}{{3 - t}}\)     M1A1

putting \(t = 1\)

\(G’(1) = 2\)     A1

\(\frac{{G''(t)G(t) - {{[G'(t)]}^2}}}{{{{[G(t)]}^2}}} =  - \frac{1}{{{t^2}}} + \frac{2}{{{{(3 - t)}^2}}}\)     M1A1

putting \(t = 1\)

\(G’’(1) - 4 =  - 1 + \frac{1}{2}\)

\(G’’(1) = \frac{7}{2}\)     A1

[??? marks]

\({\text{Var}}(X) = G''(1) + G'(1) - {[G'(1)]^2} = \frac{3}{2}\)     A1

[??? marks]