Answer :
[tex]p(x)=\begin{cases}\dfrac1n&\text{for }x\in\{1,2,\ldots,n\}\\\\0&\text{otherwise}\end{cases}[/tex]
describes a discrete uniform distribution. The expectation is given by
[tex]\mathbb E(X)=\displaystyle\sum_xxp(x)=\sum_{x=1}^nxp(x)=\sum_{x=1}^n\dfrac xn[/tex]
Since [tex]\dfrac1n[/tex] is independent of [tex]x[/tex], you have
[tex]\displaystyle\sum_{x=1}^n\frac xn=\frac1n\sum_{x=1}^nx=\frac{\dfrac{n(n+1)}2}n=\dfrac{n+1}2[/tex]
describes a discrete uniform distribution. The expectation is given by
[tex]\mathbb E(X)=\displaystyle\sum_xxp(x)=\sum_{x=1}^nxp(x)=\sum_{x=1}^n\dfrac xn[/tex]
Since [tex]\dfrac1n[/tex] is independent of [tex]x[/tex], you have
[tex]\displaystyle\sum_{x=1}^n\frac xn=\frac1n\sum_{x=1}^nx=\frac{\dfrac{n(n+1)}2}n=\dfrac{n+1}2[/tex]