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
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\begin{align} E(Y) &= \dfrac{1}{n}\sum_{k=1}^{n} \left[ k \left( \dfrac{1}{n} + \dfrac{1}{n-1} + \dots + \dfrac{1}{n – k + 1} \right) \right] \ &= \dfrac{1}{n} \left{ 1 \times \dfrac{1}{n} + 2 \left( \dfrac{1}{n} + \dfrac{1}{n-1} \right) + \dots + k \left( \dfrac{1}{n} + \dfrac{1}{n-1} + \dots + \dfrac{1}{1} \right) \right} \ &= \dfrac{1}{n} \left( \dfrac{1+2+\dots+n}{n} + \dfrac{2+\dots+n}{n-1} + \dots + \dfrac{n}{1} \right) \ &= \dfrac{1}{n} \left( \dfrac{n+1}{2} + \dfrac{n+2}{2} + \dots + \dfrac{n+n}{2} \right) \ &= \dfrac{(3n+1)n}{2n \times 2} = \dfrac{3n+1}{4}. \end{align}
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