At first I had this summation :
$$ \sum_{m=1}^{n} (m)\binom{n}{m}(3^m -2^m) $$
After that I used the identity $\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}$
Which gave me the modified form of the summation: $$n\sum_{m=2}^{n-1}\binom{n-1}{m-1}(3^m-2^m)$$
But when I enter $n=2$ and $n=3$ , the two forms of the summation give different answers.
Your use of the quoted identity should lead you to rewrite the sum as $n\sum_{m=\color{blue}{1}}^\color{blue}{n}\binom{n-1}{m-1}(3^m-2^m)$. My guess is you thought $\binom{n-1}{0}=\binom{n-1}{n-1}=0$, which would allow you to delete two terms as in your limits, but in act $\binom{n-1}{0}=\binom{n-1}{n-1}=1$.