A 6-sided die is weighted so that the probability of any number being rolled is proportional to the value of the roll.  (So, for example, the probability of a 2 being rolled is twice that of a 1 being rolled.)  What is the expected value of a roll of this weighted die?  Express your answer as a common fraction.
Let $x$ be the probability that a 1 is rolled.  Then the probability that a 2 is rolled is $2x$, the probability that a 3 is rolled is $3x$, and so on.  Since the sum of all these probabilities must be 1, we have that $x + 2x + \cdots + 6x = 1$, which means that $21x = 1$, so $x = \frac{1}{21}$.  Therefore \[ E = \frac{1}{21}(1) + \frac{2}{21}(2) + \cdots + \frac{6}{21}(6) = \frac{1^2 + 2^2 + \cdots 6^2}{21} = \boxed{\frac{13}{3}}. \]