\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Since \[ S = \sum _{n = 0}^{\infty } (1 + nd) r^n, \] we have \begin {align*} (1 - r)S & = S - rS \\ & = \sum _{n = 0}^{\infty } (1 + nd) r^n - \sum _{n = 0}^{\infty } (1 + nd) r^{n + 1} \\ & = \sum _{n = 0}^{\infty } (1 + nd) r^n - \sum _{n = 1}^{\infty } (1 + (n - 1)d) r^n \\ & = 1 + \sum _{n = 1}^{\infty } d r^n \\ & = 1 + \frac {dr}{1 - r}, \end {align*}
and hence \[ S = \frac {1}{1 - r} + \frac {dr}{(1 - r)^2}, \] as desired.
Let \(X\) be the number shots taken for Arthur to hit the target for the first time, and we have \(X \sim \Geometric (a)\), we would like to show \(\Expt (X) = \frac {1}{a}\).
The probability mass function for \(X\) satisfies \[ \Prob (X = x) = (1 - a)^{x - 1} \cdot a, \] and hence \begin {align*} \Expt (X) & = \sum _{x = 1}^{\infty } x \Prob (X = x) \\ & = a \cdot \sum _{x = 1}^{\infty } x (1 - a)^{x - 1} \\ & = a \cdot \sum _{x = 0}^{\infty } (1 + x) (1 - a)^{x} \\ & = a \cdot \left [\frac {1}{1 - (1 - a)} + \frac {1 \cdot (1 - a)}{(1 - (1 - a))^2}\right ] \\ & = a \cdot \left [\frac {1}{a} + \frac {1 - a}{a^2}\right ] \\ & = a \cdot \frac {1}{a^2} \\ & = \frac {1}{a}, \end {align*}
as desired.
Since there is a probability \(a\) of Arthur winning on a particular shot, and if Arthur did not hit (with probability \((1 - a)\)), then there is a probability \(b\) of Boadicea winning on the shot, and \((1 - b)\) probability that the first two shots both miss, and the game continues as if nothing happened in the first two shots. Therefore, \[ (\alpha , \beta ) = a (1, 0) + (1 - a) b (0, 1) + (1 - a)(1 - b) (\alpha , \beta ), \] and hence \[ \left \{ \begin {aligned} \alpha & = a + (1 - a)(1 - b) \alpha = a + a' b' \alpha , \\ \beta & = (1 - a) b + (1 - a)(1 - b) \beta = a' + a' b' \beta , \end {aligned} \right . \implies \left \{ \begin {aligned} \alpha & = \frac {a}{1 - a' b'}, \\ \beta & = \frac {a' b}{1 - a' b'}. \end {aligned} \right . \]
Let the expected number of shots in the contest be \(e\). By the linearity of the expectation, we have \[ e = a \cdot 1 + a' b \cdot 2 + a' b' \cdot (e + 2), \] where the \((e + 2)\) comes from when Arthur and Boadicea both miss their initial shots (for the \(2\)), and the game continues (for the \(e\)), and hence \[ e = \frac {a + 2a'b + 2a' b'}{1 - a' b'} = \frac {a + 2a'}{1 - a' b'} = \frac {2 - a}{1 - a' b'}. \]
On the other hand, we have \[ \frac {\alpha }{a} + \frac {\beta }{b} = \frac {1}{1 - a' b'} + \frac {1 - a}{1 - a' b'} = \frac {2 - a}{1 - a' b'}, \] and therefore \[ e = \frac {\alpha }{a} + \frac {\beta }{b}, \] as desired.