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The p.m.f. of a random variable $\mathrm{X}$ is $\mathrm{P}(x)=\left\{\begin{array}{cl}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)} & , \quad x=1,2,3, \ldots \mathrm{n} \\ 0 & , \text { otherwise }\end{array}\right.$, then $\mathrm{E}(\mathrm{X})$ is
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The correct answer is:
$\frac{\mathrm{2n}+1}{3}$
\begin{array}{|l|c|c|c|c|c|}
\hlineX & 1 & 2 & 3 & \ldots & n \\
\hlineP(X) & \frac{2}{n(n+1)} & \frac{4}{n(n+1)} & \frac{6}{n(n+1)} & \cdots & \frac{2 n}{n(n+1)} \\
\hline
\end{array}
$\begin{aligned} \mathrm{E}(\mathrm{X}) & =\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \\ & =1 \cdot \frac{2}{\mathrm{n}(\mathrm{n}+1)}+2 \cdot \frac{4}{\mathrm{n}(\mathrm{n}+1)}+3 \cdot \frac{6}{\mathrm{n}(\mathrm{n}+1)} \\ & +\ldots+\mathrm{n} \cdot \frac{2 \mathrm{n}}{\mathrm{n}(\mathrm{n}+1)} \\ & =\frac{2}{\mathrm{n}(\mathrm{n}+1)}\left(1+4+9+\ldots+\mathrm{n}^2\right) \\ & =\frac{2}{\mathrm{n}(\mathrm{n}+1)}\left(1^2+2^2+3^2+\ldots+\mathrm{n}^2\right) \\ & =\frac{2}{\mathrm{n}(\mathrm{n}+1)} \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6} \\ & =\frac{2 \mathrm{n}+1}{3}\end{aligned}$
\hlineX & 1 & 2 & 3 & \ldots & n \\
\hlineP(X) & \frac{2}{n(n+1)} & \frac{4}{n(n+1)} & \frac{6}{n(n+1)} & \cdots & \frac{2 n}{n(n+1)} \\
\hline
\end{array}
$\begin{aligned} \mathrm{E}(\mathrm{X}) & =\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \\ & =1 \cdot \frac{2}{\mathrm{n}(\mathrm{n}+1)}+2 \cdot \frac{4}{\mathrm{n}(\mathrm{n}+1)}+3 \cdot \frac{6}{\mathrm{n}(\mathrm{n}+1)} \\ & +\ldots+\mathrm{n} \cdot \frac{2 \mathrm{n}}{\mathrm{n}(\mathrm{n}+1)} \\ & =\frac{2}{\mathrm{n}(\mathrm{n}+1)}\left(1+4+9+\ldots+\mathrm{n}^2\right) \\ & =\frac{2}{\mathrm{n}(\mathrm{n}+1)}\left(1^2+2^2+3^2+\ldots+\mathrm{n}^2\right) \\ & =\frac{2}{\mathrm{n}(\mathrm{n}+1)} \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6} \\ & =\frac{2 \mathrm{n}+1}{3}\end{aligned}$
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