Vibe, check! Does this problem give you the vibe that it might be telescoped? We hope so! In the absence of any bright idea, let us try to telescope this sum. That we want to find $V_{r}$ such that $T_{r}=V_{r}-V_{r-1}$ or $V_{r-1}-V_{r}.$ Look at the denominator, it has something like $r(r+1) $. So, it is already in the given form. But there is a nagging factor of $2$ in the denominator! So, if we can find a sequence $u_{r}$ such that
$$ T_{r} = \frac{2r^2-r^3+2r-2}{2^r r(r+1)} = \frac{2 (r+1) u_{r}- r u_{r+1}}{2^r r(r+1)} $$
then we are done. Notice, given all clues $u_{r}$ must be a polynomial. So, we have $ 2r^2-r^3+2r-2 = 2 (r+1) u_{r}- r u_{r+1}. $ Therefore, let $u_{r}=ar^2 + br+ c$ and compare the coefficients on both sides, we get: $u_{r}=(r-1)^2$
$$ \implies T_{r} = \frac{2r^2-r^3+2r-2}{2^r r(r+1)} = \frac{2(r+1)(r-1)^2-r (r)^2}{2^r r(r+1)} $$
$$ \implies T_{r} = \frac{(r-1)^2}{2^{r-1}r}- \frac{r^2}{2^r(r+1)} $$
$$ \implies \sum_{r=1}^{n} T_{r} = \sum_{r=1}^n \left ( \frac{(r-1)^2}{2^{r-1}r}- \frac{r^2}{2^r(r+1)} \right) = - \frac{n^2}{2^n(n+1)}. $$
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