Asked by Mikal
solve for r
r^64-1 = 2.5x10^7 (r-1)
any ideas?
r^64-1 = 2.5x10^7 (r-1)
any ideas?
Answers
Answered by
Reiny
r-1 is a factor of r^64 - 1 so use synthetic division (notice the pattern) to divide both sides by r-1 to get
r^63 + r^62 + ... + r + 1 = 2.5(10)^7
the left side is a geometric series with
a = 1
common ratio = r
n = 64
so the sum of the left side = 1(r^64 - 1)/63
then (r^63 - 1)/63 = 2.5(10^7
r^63 = 63(2.5)(10)^7 + 1
taking the 63rd root I got
r = 1.3922
r^63 + r^62 + ... + r + 1 = 2.5(10)^7
the left side is a geometric series with
a = 1
common ratio = r
n = 64
so the sum of the left side = 1(r^64 - 1)/63
then (r^63 - 1)/63 = 2.5(10^7
r^63 = 63(2.5)(10)^7 + 1
taking the 63rd root I got
r = 1.3922
Answered by
Reiny
OOOPS, just looked at my solution again and
I'm WRONG!!! (must have had an early morning synapse collapse)
The sum of my series should have said
1(r^64 - 1)/(r-1)
then r^64 - 1 = 2.5(10)^7(r-1)
which brings me back to what we started with.
Arggh!!!
Sorry, looks like we would have to use some method such as Newtons's Method or some kind of "equation solver" program.
I'm WRONG!!! (must have had an early morning synapse collapse)
The sum of my series should have said
1(r^64 - 1)/(r-1)
then r^64 - 1 = 2.5(10)^7(r-1)
which brings me back to what we started with.
Arggh!!!
Sorry, looks like we would have to use some method such as Newtons's Method or some kind of "equation solver" program.
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