Question

find the sum for n = 7, a = 2, and r = 1/2.

I get 127/128...but the answers say 127/32.

Answer 1

I use the formula Sn = (1 - 2^n) / 1-r

Answer 2

sorry a(1 - 2^n) / 1-r

Answer 3

you are right

Answer 4

Really ? Thanks

Answer 5

your welcome

Answer 6

confused

Answer 7

\[r=\frac{1}{2}\]?

Answer 8

MY BAD !

a(1 - ****r^n***) / 1-r

Answer 9

n = 7, a = 2, and r = 1/2.

Answer 10

r is common ratio

Answer 11

i am assuming this is
\[a=2,r=\frac{1}{2}, n = 7\] in other words you have
\[2+1+\frac{1}{2}+\frac{4}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\]

Answer 12

yep

Answer 13

typos above, should be
\[\frac{1}{4}\]

Answer 14

Yeah O got that :)

Answer 15

*I

Answer 16

so I did s7 = 2(1 - (1/2)^7)/1 - 1/2

Answer 17

shouldbe
\[S_n=\frac{a-ar^{n-1}}{1-r}\]

Answer 18

off by a power

Answer 19

damn another typo, should be
\[S_n=\frac{a-ar^{n+1}}{1-r}\]

Answer 20

I my formula has a factored out at the top, same thingg I think so

Answer 21

so you have
\[\frac{2-(\frac{1}{2})^8}{1-\frac{1}{2}}\]

Answer 22

oh okay

Answer 23

btw in any case the first to numbers you are adding are 2 and 1, so there is no way you can get
\[\frac{127}{128}\] right?

Answer 24

d'awww

Answer 25

sum has to be greater than 3 for sure

Answer 26

yup true

Answer 27

point is not to call you out, but to remind you not to get married to a formula and forget about common sense

Answer 28

You're right, haha...thanks

Answer 29

OMG! I know what I did wrong now! I divided a fraction by a fraction wrong (always do that ugh!)....

Answer 30

Thanks for your help though ~~~~ Otherwise I would have assumed the paper was wrong :P