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counting and probability intermediate

Problem

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.
Solution
Let be the probability that a 1 is rolled. Then the probability that a 2 is rolled is , the probability that a 3 is rolled is , and so on. Since the sum of all these probabilities must be 1, we have that , which means that , so . Therefore
Final answer
\frac{13}{3}