\(% 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}}\)
First, note that \begin {align*} 1 & = \sum _{x, y = 1}^{x = n} \Prob (X = x, Y = y) \\ & = \sum _{x = 1}^{n} \sum _{y = 1}^{n} k(x + y) \\ & = \sum _{x = 1}^{n} \sum _{y = 1}^{n} (kx + ky) \\ & = \sum _{x = 1}^{n} \left (n \cdot kx + k\sum _{y = 1}^{n} y\right ) \\ & = nk\sum _{x = 1}^{n} x + nk\sum _{y = 1}^{n} y \\ & = n^2 (n+1) k \end {align*}
Therefore, \(k = \frac {1}{n^2(n+1)}\)
\begin {align*} \Prob (X = x) & = \sum _{y = 1}^{n} \Prob (X = x, Y = y) \\ & = \sum _{y = 1}^{n} k(x + y) \\ & = nkx + k\sum _{y = 1}^{n} y \\ & = nkx + \frac {kn(n+1)}{2} \\ & = \frac {x}{n(n+1)} + \frac {1}{2n} \\ & = \frac {2x + n+1}{2n(n+1)}, \end {align*}
as desired.
By symmetry, \(\Prob (Y = y) = \frac {2y + n+1}{2n(n+1)}\).
We have \[ \Prob (X = x) \cdot \Prob (Y = y) = \frac {(2x+n+1)(2y+n+1)}{4n^2(n+1)^2}. \]
But \(\Prob (X = x, Y = y) = \frac {x+y}{n^2(n+1)}\) is not equal to this. So \(X\) and \(Y\) are not independent.
By definition, \[ \Cov (X, Y) = \Expt (XY) - \Expt (X)\Expt (Y). \]
We have \begin {align*} \Expt (X) = \Expt (Y) & = \sum _{t = 1}^{n} t \cdot \Prob (X = t) \\ & = \sum _{t = 1}^{n} \frac {t \cdot (2t + n + 1)}{2n(n+1)} \\ & = \frac {1}{n(n+1)} \sum _{t = 1}^{n} t^2 + \frac {1}{2n} \sum _{t = 1}^{n} t \\ & = \frac {n(n+1)(2n+1)}{6n(n+1)} + \frac {n(n+1)}{4n} \\ & = \frac {2n+1}{6} + \frac {n+1}{4} \\ & = \frac {4n+2+3n+3}{12} \\ & = \frac {7n+5}{12}, \end {align*}
and \begin {align*} \Expt (XY) & = \sum _{x, y = 1}^{n} xy \cdot \Prob (X = x, Y = y) \\ & = \sum _{x = 1}^{n} \sum _{y = 1}^{n} \frac {xy(x + y)}{n^2(n+1)} \\ & = \frac {1}{n^2(n+1)} \sum _{x = 1}^{n} \sum _{y = 1}^{n} xy(x + y) \\ & = \frac {1}{n^2(n+1)} \sum _{x = 1}^{n} \sum _{y = 1}^{n} (x^2y + xy^2) \\ & = \frac {1}{n^2(n+1)} \left [\sum _{x = 1}^{n} x^2 \sum _{y = 1}^{n} y + \sum _{x = 1}^{n} x \sum _{y = 1}^{n} y^2\right ] \\ & = \frac {1}{n^2(n+1)} \cdot 2 \cdot \frac {n(n+1)(2n+1)}{6} \cdot \frac {n(n+1)}{2} \\ & = \frac {(2n+1)(n+1)}{6}. \end {align*}
Therefore, \begin {align*} \Cov (X, Y) & = \Expt (XY) - \Expt (X)\Expt (Y) \\ & = \frac {(2n+1)(n+1)}{6} - \frac {(7n+5)^2}{144} \\ & = \frac {48n^2 + 72n + 24}{144} - \frac {49n^2 + 70n + 25}{144} \\ & = \frac {-n^2 + 2n - 1}{144} \\ & = -\frac {(n-1)^2}{144} \\ & < 0, \end {align*}
as desired.