in general $\sum \limits_{k=0}^n \alpha^k = \dfrac{1-\alpha^{n+1}}{1-\alpha}$

For the second one

$\sum \limits_{k=3}^n (-1)^{k+1}2^k = \\

\sum \limits_{k=0}^{n-3} (-1)^{k+4}2^{k+3} = \\

\dfrac 1 2 \sum \limits_{k=0}^{n-3}(-2)^{k+4} = \\

8\sum \limits_{k=0}^{n-3} (-2)^k = \\

8 \dfrac{1-(-2)^{n-2}}{1-(-2)} = \\

\dfrac 8 3 \cdot \left(1 - (-2)^{n-2}\right)

$