Question

PLEASE HURRY 50 POINTS Answer Both parts and I will reward Brainliest

Answer 1

We have:

$a_1=-3$

and:

$r=4$

so:

$\sum\limits_{n=2}^5\,-3\cdot4^{n-1}=\sum\limits_{n=1}^5\,-3\cdot4^{n-1}-a_1=S_5-a_1=\frac{-3-(-3)\cdot 4^5}{1-4}-(-3)=$

$=\dfrac{-3-(-3)\cdot 1024}{-3}+3=\dfrac{-3(1-1024)}{-3}+3=1-1024+3=\boxed{-1020}$

It is a finite sum, so it is convergent.