25~26高三下·山东日照一模·第14题

$\dfrac{3n+1}{4}$

$Y$可能的取值为$1,2,\dots,n$，由全概率公式有：
${P(Y = k)}$${ = \sum\limits_{i = 1}^k P (Y = k|X = i)P(X = i)}$
${ = \frac{1}{n}\sum\limits_{i = 1}^k {\frac{1}{{n – i + 1}}} }$${ = \frac{1}{n}\left( {\frac{1}{n} + \frac{1}{{n – 1}} + \ldots + \frac{1}{{n – k + 1}}} \right)}$，
故$E(Y) = \frac{1}{n}\sum\limits_{k = 1}^n {\left[ {k\left( {\frac{1}{n} + \frac{1}{{n – 1}} + \ldots + \frac{1}{{n – k + 1}}} \right)} \right]} \; $
$= \frac{1}{n}${$\rm{{ }}1 \times \frac{1}{n} + 2\left( {\frac{1}{n} + \frac{1}{{n – 1}}} \right) + \ldots + k\left( {\frac{1}{n} + \frac{1}{{n – 1}} + \ldots + \frac{1}{1}} \right)$}
$= \frac{1}{n}\left( {\frac{{1 + 2 + \ldots + n}}{n} + \frac{{2 + \ldots + n}}{{n – 1}} + \ldots + \frac{n}{1}} \right)\;$
$= \frac{1}{n}\left( {\frac{{n + 1}}{2} + \frac{{n + 2}}{2} + \ldots + \frac{{n + n}}{2}} \right)\; = \frac{{(3n + 1)n}}{{2n \times 2}} = \frac{{3n + 1}}{4}
$