Question

can someone please help with the steps geometric series? PLEASE!!!

Answer 1

I understan there are 8 terms yes? and that i =1. I think I add and mutiply but I'm not getting the right answer so I guess I'm not doing the steps right.

Answer 2

Sum for geometric series is
\(\Large S_n = a_1 \dfrac{r^{n-1}}{r-1}\)

first term = a1 = 2^0 = 1
n= number of terms = 8
you can find 'r' ?

Answer 3

you tried adding 8 terms ?
what would you have done if there were 100000 terms? :P

Answer 4

2i=s^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8 Is this correct??

Answer 5

the exponent is i-1
and i starts from 1

the first term will be 2^(1-1) = 2^0
right ?

Answer 6

so your sum will be
S = 2^0 +2^1 + ...+2^7

but please don't do it this way

Answer 7

This is the way my lesson said to and the answer I got was 510.

Answer 8

ok let me learn your way

Answer 9

so I need to find for r?

Answer 10

if your lesson specifically asked then find :
2^0 +2^1 + ...+2^7

(you got why 2^1 + ...+2^8 is incorrect?)

Answer 11

r is the common ratio :
r = 2nd term/1st term = 3rd term/2nd term = ...and so on

Answer 12

so I did it incorrect?

Answer 13

the common ratio is 2

Answer 14

yes and yes

Answer 15

r =2,
now just use the formula!

Answer 16

also verify by doing
2^0 +2^1 + ...+2^7

Answer 17

:( this class is going to kill me. ok give me a sec

Answer 18

thanks for helping me

Answer 19

happy to help ^_^
let me know once you get the answer, i'll verify it :)

Answer 20

so sorry for the typo
\(\Large S_n = a_1 \dfrac{r^{n}-1}{r-1}\)

Answer 21

sorry confused as to what a is and Sn

Answer 22

Sn is the sum of n terms, the answer that u need

a1 is the first term of the series

for your series

2^0, 2^1, ...,2^7

first term = 2^0 =1

Answer 23

What do you call that goofy E?

Answer 24

what goofy E ?

Answer 25

In the equation

Answer 26

My lesson doesn't even say what it is called or mean

Answer 27

oh
\(\sum \)

thats the sign of summation

Answer 28

\(\sum_{i=1}^{8} 2^{i-1}\)

means add(sum) all the terms starting from i=1 in 2^{i-1} till i =8

Answer 29

'add(sum) all the terms starting from i=1 in 2^{i-1} till i =8'

= 2^{1-1} + 2^{2-1} + ... 2^{8-1}

Answer 30

= 2^0 +2^1+ ...2^7

Answer 31

= 1+2+4+8+16+32+64+128

Answer 32

I got 251 but that isn't one of my answers

Answer 33

how 251?

Answer 34

\(\Large S_n = a_1 \dfrac{r^{n}-1}{r-1} = \dfrac{2^8 -1}{2-1} = 2^8-1 = ..?\)

Answer 35

omg im insane Iyou raised each one to the power and then add

Answer 36

now you got it :)

Answer 37

255

Answer 38

\(\huge \checkmark \)

Answer 39

1+2+4+8+16+32+64+128 =255

Answer 40

2^8-1 = 255

Answer 41

im so sorry thanks so much for sticking it out. can I call on you again. Is this even algebra 2?

Answer 42

welcome ^_^
sure you can ask me for any help again :)
not sure what algebra2 in your country covers...