Question

I suck at these

Answer 1

at what

Answer 2

ice cream? me too, I like sucking them too

Answer 3

lol

Answer 4

I love ice cream

Answer 5

come on guys focus

Answer 6

one sec while I write it

Answer 7

\[\sum_{i=1}^{infinity} 7(4)^{i-1}\]

Answer 8

Find the 6th partial sum of

Answer 9

one sec

Answer 10

\(\large {\huge \Sigma}_{i=1}^{\infty}\quad a_1{\color{blue}{ r}}^{i-1}\implies a_1\left(\cfrac{1-{\color{blue}{ r}}^i}{i-{\color{blue}{ r}}}\right)\)_

Answer 11

hmmm lemme fix that

Answer 12

Ok

Answer 13

\(\large {\huge \Sigma}_{i=1}^{\infty}\quad a_1{\color{blue}{ r}}^{i-1}\implies a_1\left(\cfrac{1-{\color{blue}{ r}}^i}{1-{\color{blue}{ r}}}\right)\)

Answer 14

notice, you're asked for the "6th partial sum" thus \(\large \bf {\huge \Sigma}_{i=1}^{\infty}\quad 7{\color{blue}{ 4}}^{{\color{brown}{ i}}-1}\implies 7\left(\cfrac{1-{\color{blue}{ 4}}^{\color{brown}{ 6}}}{1-{\color{blue}{ 4}}}\right)\)

Answer 15

alright

Answer 16

hmmm I should separate the 7 and 4...so \(\Large \bf {\huge \Sigma}_{i=1}^{\infty}\quad 7{\color{blue}{ (4)}}^{{\color{brown}{ i}}-1}\implies
7\left(\cfrac{1-{\color{blue}{ 4}}^{\color{brown}{ 6}}}{1-{\color{blue}{ 4}}}\right)\)

Answer 17

but anyhow... that's how you'd get the 6th partial sum

Answer 18

so 7(85)?

Answer 19

@jdoe0001

Answer 20

hmmm not quite

Answer 21

recall your PEMDAS :)

Answer 22

ok what should I do?

Answer 23

well... you'd do the subtractions in the fraction first
then
the multiplication, and then the division

Answer 24

\(\bf 7\left(\cfrac{1-4^6}{1-4}\right)\implies \cfrac{7(1-4^6)}{1-4}\)

Answer 25

595?

Answer 26

hmmm.... check your operations...nope,, is a bigger value

Answer 27

but the 6th root of -4 is -256

Answer 28

oh wait its 256

Answer 29

im so lost lol

Answer 30

oh wait its 9555