Calculate $$\sum_{n=0}^{17} 4\cdot(-2)^n$$
See development and solution
Development:
Note that the first term that this is added is $$n=0$$, so, as we know, in order to be able to do the sum we change $$m$$ with $$m=n+1$$, so we have:
$$$\left. \begin{array}{c} n=m-1 \\ n=17 \Rightarrow m=18 \\ n=0 \Rightarrow m=1 \end{array} \right\} \Rightarrow \sum_{n=0}^{17} 4\cdot(-2)^n = \sum_{m=1}^{18} 4\cdot(-2)^{m-1}$$$
And in this way we find that $$$\sum_{m=1}^{18} 4\cdot(-2)^{m-1}=\dfrac{4(1-(-2)^{18})}{1-(-2)}=\dfrac{4+2^{18}}{3}=\dfrac{262.148}{3}$$$
Solution:
$$\sum_{n=0}^{17} 4\cdot(-2)^n =\dfrac{262.148}{3}$$