Calculate series sum(turning bounds)? $\sum_{n+1}^0 -k(3^k) $
assuming (n+1)<0
As far as I understand I should use the formula for $\sum_{0}^{N-1} k(a^k) $ but what should I do the get to this formula?

$a+b = b+a$. It doesn't matter which order you add  $$\sum_{k=n+1}^{k=0} -k(3^k) = \sum_{k=0}^{k=n+1} -k(3^k) $$
Next let $u = -k$,
when $k=0, u = 0$
when $k=n+1, u = -(n+1)$ 
$$ \sum_{k=0}^{k=n+1} -k(3^k) =  \sum_{u=0}^{u=-(n+1)} u(3^{-u}) = \sum_{u=0}^{u=-(n+1)} u(1/3)^u   $$
