Expectation value of n dots

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I am current doing this question and my answer compared to the numerical solution does not match up but I can see what I am doing wrong.

Question:

The number of dots showing on a die is $n$ and $f(n)$ is some function of $n$. If you were to roll many many dice, would would be the mean value of $f$ for $$f=(n+2)^2$$

My answer is a follows:

$$\bar{f}=\sum \left(n+2\right)^2P$$

Summing over the six sides of the die with $P=1/6$ for each side gives us

$$\bar{f}=\frac{1}{6}\sum_{n=1}^6 \left(n+2\right)^2=\frac{199}{2}=33.5$$

However the book make it $33\frac{1}{6}$. Which I can see being correct as the only this value is produce is if $$ \bar{f}=\sum \left(n+2\right)P. $$ Is this a typo in the book or have I missed understood something in the question?

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You make one slight error: $$\bar{f}=\sum\left(n+2\right)^2 \cdot P=\frac{1}{6}\sum_{n=1}^6 \left(n+2\right)^2=\frac{199}{\mathbf{6}}$$