Second generalized Bernoulli number $B_{2,\chi}$

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Let be $\chi$ (non-trivial) Dirichlet charakter of conductor $f$. Then I know that $$B_{n,\chi}=f^{n-1}\sum_{a=1}^f\chi(a)B_n\left(\frac{a}{f}\right).$$ Assume $\chi(-1)=1$ and plug $n=2,$ then $$B_{2,\chi}=f\sum_{a=1}^f\chi(a)\left(\frac{a^2}{f^2}-\frac{a}{f}+\frac{1}{6}\right).$$ Using $\sum_{a=1}^f\chi(a)=0$ I get $$B_{2,\chi}=\frac{1}{f}\sum_{a=1}^f\chi(a)a^2-\sum_{a=1}^f\chi(a)a.$$ I would like to show that the second sum equals to $0,$ but I don't know how to do it. I haven't used the assumption $\chi(-1)=1$ yet, but I don't see why it is helpful.

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$$\sum_{a=1}^f\chi(a)a=\sum_{a=1}^{f-1}\chi(a)a=\sum_{a=1}^{f-1}(f-a)\chi(f-a)=\sum_{a=1}^{f-1}(f-a)\chi(a)$$ $$\Rightarrow2\sum_{a=1}^{f-1}\chi(a)a=f\sum_{a=1}^{f-1}\chi(a)=0$$