\(% 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}}\)
By the formula of the sum for a geometric series, we have \begin {align*} \sum _{r = 0}^{n - 1} \exp (2i (\alpha + r\pi / n)) & = \exp (2i (\alpha + 0\pi / n)) \cdot \frac {1 - \exp (2 i \pi / n)^ n}{1 - \exp (2 i \pi / n)} \\ & = \exp (2 i \alpha ) \cdot \frac {1 - \exp (2 i \pi )}{1 - \exp (2 i \pi / n)} \\ & = \exp (2 i \alpha ) \cdot \frac {1 - 1}{1 - \exp (2 i \pi / n)} \\ & = 0, \end {align*}
since the denominator is not \(0\).
By geometry, we have \[ r \cos \theta + s = d, \] and hence \[ s = d - r \cos \theta . \]
Since \(r = ks = k(d - r \cos \theta )\), we have \[ r = \frac {kd}{1 + k \cos \theta }. \]
Let \(L_1\) be an angle \(\alpha \) to horizontal, then \(L_j\) is angle \(\alpha + (j - 1)\pi / n\) angle to the horizontal for \(j = 1, 2, \ldots , n\). Let \(\theta _j = \alpha + (j - 1)\pi / n\), and we have \begin {align*} l_j & = \LEvalAt {r}{\theta = \theta _j} + \LEvalAt {r}{\theta = \theta _j + \pi } \\ & = kd \left (\frac {1}{1 + k \cos \theta _j} + \frac {1}{1 + k \cos \left (\theta _j + \pi \right )}\right ) \\ & = kd \left (\frac {1}{1 + k \cos \theta _j} + \frac {1}{1 - k \cos \theta _j}\right ) \\ & = kd \cdot \frac {1 + k \cos \theta _j + 1 - k \cos \theta _j}{1 - k^2 \cos ^2 \theta _j} \\ & = \frac {2kd}{1 - k^2 \cos ^2 \theta _j}. \end {align*}
Hence, we have \begin {align*} \sum _{j = 1}^{n} \frac {1}{l_j} & = \frac {1}{2kd} \sum _{j = 1}^{n}(1 - k^2 \cos ^2 \theta _j) \\ & = \frac {1}{2kd} \left [n - k^2 \sum _{j = 1}^{n} \cos ^2 \left (\alpha + (j - 1)\pi / n\right )\right ] \\ & = \frac {1}{2kd} \left [n - \frac {k^2}{2} \cdot \sum _{j = 1}^{n} \left [1 + \cos 2 \left (\alpha + (j - 1)\pi / n\right )\right ]\right ] \\ & = \frac {1}{2kd} \left [n - \frac {nk^2}{2} - \frac {k^2}{2}\cdot \sum _{j = 1}^{n} \cos 2 \left (\alpha + (j - 1)\pi / n\right )\right ] \\ & = \frac {1}{2kd} \left [n - \frac {nk^2}{2} - \frac {k^2}{2}\cdot \sum _{r = 0}^{n - 1} \cos 2 \left (\alpha + r\pi / n\right )\right ] \\ & = \frac {1}{2kd} \left [n - \frac {nk^2}{2} - \frac {k^2}{2}\cdot \sum _{r = 0}^{n - 1} \re \exp (2i \left (\alpha + r\pi / n\right ))\right ] \\ & = \frac {1}{2kd} \left [n - \frac {nk^2}{2} - \frac {k^2}{2}\cdot 0\right ] \\ & = \frac {1}{2kd} \cdot \frac {n (2 - k^2)}{2} \\ & = \frac {n (2 - k^2)}{4kd}, \end {align*}
as desired.