chassicat
chassicat chassicat
  • 13-08-2018
  • Mathematics
contestada

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

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

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WojtekR
WojtekR WojtekR
  • 14-08-2018

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.

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