Summation problem (not sure of my solution)

Question

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?

Answer 1

That's Wrong.

If $f(x) = 3-x$ then $f(2x) \neq 2(3-x)$ but rather $f(2x) = 3-2x$.

So if you have some function $g(x)$ as the input of $f$ , then you need to replace all the $x$ in $f(x)$ into $g(x)$

Answer 2

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*}