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Mathematics 20 Online
OpenStudy (anonymous):

how do you find the Geometric sum of 24+12+6+3 . . .

OpenStudy (anonymous):

If a and b are positive integers such that ab = 125, then (a-b)^(a+b-4) is equal to ?

OpenStudy (anonymous):

I don't know what that means haha im super confused

OpenStudy (anonymous):

Rule of the sum:\[24*\sum_{0}^{\infty} 1/2^n\] Nifty trick for geometric series with initial number a and ratio r:\[24*\sum_{0}^{\infty} 1/2^n = a/(1-r) = 24/(1-1/2) = 48\]

OpenStudy (anonymous):

well you got the right answer. i just don't understand it

OpenStudy (anonymous):

The trick is: \[a*\sum_{0}^{\infty}(r)^n = a/(1-r)\] As long as -1 < r < 1

OpenStudy (anonymous):

i used S = t1 / 1-r

OpenStudy (anonymous):

but i got 12

OpenStudy (anonymous):

That's the same thing :D

OpenStudy (anonymous):

then how did I get 12? hahaha

OpenStudy (anonymous):

You multiplied by 1/2 instead of divided 24/(1/2) = 48, not 12

OpenStudy (anonymous):

OHHHHH

OpenStudy (anonymous):

wow that was embarassing hahaha

OpenStudy (anonymous):

thanks daniel :)

OpenStudy (anonymous):

No it isn't. I do that all the time :D

OpenStudy (anonymous):

No worries :D

OpenStudy (anonymous):

i actually need a lot more help

OpenStudy (anonymous):

if youve got some extra time :D

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