Compute $\sum_{k=-1}^{n=24}C(25,k+2)k2^k$ 
Compute $\sum_{k=-1}^{n=24}C(25,k+2)k2^k$

Well, I've found a solution for it, but I don't understand the line in the orange rectangle, can anyone exlain it please?


$$\begin{align*}
\sum_{t=1}^{26}\binom{25}t(t-2)2^{t-2}&=\sum_{t=1}^{26}\left(\binom{25}tt2^{t-2}-\binom{25}t2\cdot2^{t-2}\right)\\
&=\sum_{t=1}^{26}\binom{25}tt2^{t-2}-\sum_{t=1}^{26}\binom{25}t2^{t-1}\\
&=\sum_{t=1}^{26}\left(\binom{25}tt2^{t-2}\cdot 2\cdot\frac12\right)-\sum_{t=1}^{26}\left(\binom{25}t2^{t-1}\cdot 2\cdot\frac12\right)\\
&=\frac12\sum_{t=1}^{26}\binom{25}tt2^{t-1}-\frac12\sum_{t=1}^{26}\binom{25}t2^t\\
&=\frac12\sum_{t=1}^{25}\binom{25}tt2^{t-1}-\frac12\sum_{t=1}^{25}\binom{25}t2^t\;,
\end{align*}$$
where the last step is because $\binom{25}{26}=0$ anyway.
