expectation of $n$-sided die

Question

Assume a dice has n sides, what is the expected value of the dice?

My solution:

The PMF of the die would be 1/n, so you would multiply this by the summation of n.

Answer 1

Following definition of expected value of a discrete random variable, we have:

$$E[X_n] = \sum\limits_{x\in \Omega}x\cdot Pr(X_n=x) = \sum\limits_{k=1}^n k\cdot \frac{1}{n} = \frac{1}{n}\cdot\left(\sum\limits_{k=1}^n k\right)$$

Recognizing the sum on the right as the triangle summation, one has then:

$$=\frac{1}{n}\cdot\left(\sum\limits_{k=1}^n k\right)=\frac{1}{n}\cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$$