+0

# Sequences and Series

-2
141
2
+-8

help ;-;

May 20, 2021

#1
+544
+1

$$\sum_{k=1}^{6} -3(-4)^{k-1} = -3(-4)^{1-1} -3(-4)^{2-1}-3(-4)^{3-1}-3(-4)^{4-1}-3(-4)^{5-1}$$

$$= -3(-4)^{0} -3(-4)^{1}-3(-4)^{2}-3(-4)^{3}-3(-4)^{4}$$

$$= -3 + 12 - 48 +192 - 768$$

$$= -615$$

let me know if I did anything wrong :)

May 20, 2021
#2
+30
+1

Is this a homework problem?

If not, don't continue reading this message.

If so, but you have not read Melody's message, please do so now.

If you have done both, then what is wrong with you! I mean I have asked for help with homework, but I showed my efforts. If you tried, at least show it. I don't know whether you are struggling or not, but show that you at least tried. Otherwise You'll just be a lazy slob who expects others to do their work for him. Some people I know have tried this path and got expelled. Melody let you go easy, but seriously, block this website now. I use Blocksite, you should too. To be honest, if not the Web calc policy of no crude launguage, this would be much more graphical, don't be lazy, show some effort.

May 20, 2021
edited by fuhsaha  May 20, 2021