We have:
[tex] a_1=-3[/tex]
and:
[tex] r=4 [/tex]
so:
[tex]$\sum\limits_{n=2}^5\,-3\cdot4^{n-1}=\sum\limits_{n=1}^5\,-3\cdot4^{n-1}-a_1=S_5-a_1=\frac{-3-(-3)\cdot 4^5}{1-4}-(-3)= [/tex]
[tex] =\dfrac{-3-(-3)\cdot 1024}{-3}+3=\dfrac{-3(1-1024)}{-3}+3=1-1024+3=\boxed{-1020} [/tex]
It is a finite sum, so it is convergent.
