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Old 02-22-2011   #2
MAS1

 
Join Date: Dec 2008
Posts: 249
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Quote:
Originally Posted by Lisasmith111 View Post
Here is something I found by poking around with Peter's Series.

Question is why does it work (sort of!).

ie 1/3 - 1/9 + 1/27...

Suppose I give you a term from it say -1/6561.

Now we know the common ratio could be calculated as follows:-

term 2 / term 1 = -1/9 / 1/3 = -1/3

so common ratio = -1/3

we also know the first term = 1/3

Which means we could use the formula for finding a given term from

the series ar^(n-1)

Anywho here is the crux dear reader....

I could write -1/6561 = 1/3 * (-1/3)^(n-1)

(Just to be clear in words the RHS is a third times minus a third raised to

the power of n minus 1.)

If I solve for n using logs I get

-1 * log 1/6561 = log 1/3 + -1(n-1)log 1/3

-log 1/6561 = log 1/3 + (1-n)log 1/3

-log 1/6561 = log 1/3 + log 1/3 -nlog 1/3

(-log 1/6561 - 2log 1/3) / -log 1/3 = n = 10 (which is the wrong

answer)


However, If I do the following I get an answer which is closer to

the truth:-

So again I write -1/6561 = 1/3 * (-1/3)^(n-1)

this time I multiply through by -1 and change the power around as

follows:-

1/6561 = -1/3 * (1/3)^(1-n)

log 1/6561 = -1.log 1/3 + (1-n)log 1/3

log 1/6561 = -log 1/3 +log 1/3 -nlog 1/3

log 1/6561 = -nlog 1/3

log 1/6561 / log 1/3 = -n

n = -8

Now we can't have a negative term. So I could say we want the
absolute value of n and therefore we get n = 8. Ie -1/6561 is the
8th term in the series. Which is correct.

The trick seems to work but why does is work? (at least sort of!).

But why?

Is there a better way of doing it?
Hi Lisa:

In your attempt to solve the problem using logs, your log step is incorrect.
(-1/6561) = (1/3)(-1/3)^(n - 1)

Now you have to take the logarithm of both sides which gives:
log(-1/6561) = log[(1/3)(-1/3)^(n - 1)]
NOT -1 * log 1/6561 = log 1/3 + -1(n-1)log 1/3 which is what you have.

log(-1/6561) is not equal to -1*log(1/6561)

As you probably know taking the log of a negative number is not defined, so you cannot use that method.
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