Find $\displaystyle \sum_{r=4}^{n}(lg3^{r})$

what i did was:

$\displaystyle =(lg 3) \sum_{r=4}^{n}(r)$

$\displaystyle =(lg 3)(\frac{1}{2})(n^{2}+7n)$

i used $\displaystyle \frac{n}{2}[2a+(n-1)d]$ to expand $\displaystyle \sum_{r=4}^{n}(r)$

but the suggested answer is $\displaystyle (lg 3)(\frac{1}{2})(n-3)(n+4)$........

where did i go wrong?