Is there a fast way to find whether the order of the following is $O(T^2)$ or $O(T)$? I've been trying to find the exact thing by using the geometric series multiple time, but it is so lengthy and intricate that I make many mistakes. I wonder if there is a shortcut to this kind of problems (other than implementing it numerically!)
$$ \sum_{u=1}^T (T-u) \sum_{t=u}^T \frac{1}{2^{t-u}} $$
The exact sum is
$$\begin{align*} \sum_{u=1}^T(T-u)\sum_{t=u}^T\frac1{2^{t-u}}&=\sum_{u=1}^{T-1}(T-u)\sum_{k=0}^{T-u}\left(\frac12\right)^k\\\\ &=\sum_{v=1}^{T-1}v\sum_{k=0}^v\left(\frac12\right)^k\\\\ &=\sum_{v=1}^{T-1}2v\left(1-\frac1{2^{v+1}}\right)\\\\ &=T(T-1)-\sum_{v=1}^{T-1}\frac{v}{2^v}\\\\ &=T(T-1)-\left(2-\frac{T+1}{2^{T-1}}\right)\\\\ &=T^2-\left(1-\frac1{2^{T-1}}\right)T-\left(2-\frac1{2^{T-1}}\right)\;. \end{align*}$$