Summation problem (not sure of my solution) If $f(x)=3-x$, evaluate $\sum_{k=1}^3f(2k)$.
So this is my solution:
$f[2(3-1)]=4$
$f[2(3-2)]=2$
$f[2(3-3)]=0$
So the required sum is $4+2+0=\boxed{6}$
Is my solution correct?

With $f(x)=3-x$ we obtain
\begin{align*}
\color{blue}{\sum_{k=1}^3f(2k)}&=f(2\cdot1)+f(2\cdot 2) + f(2\cdot 3)\\
&=f(2)+f(4)+f(6)\\
&=(3-2)+(3-4)+(3-6)\\
&=1-1-3\\
&\,\,\color{blue}{=-3}
\end{align*}
