Given the explicit formula u(n) = 5^n-1, where u(1) = 1, find the recursive formula for u(n).
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OpenStudy (anonymous):
OpenStudy (xapproachesinfinity):
what did you do so far?
OpenStudy (xapproachesinfinity):
any ideas
OpenStudy (anonymous):
nothing. I have no idea how to do this
OpenStudy (xapproachesinfinity):
how about finding u2 u3 and see what is going on here
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OpenStudy (xapproachesinfinity):
can you do u2
OpenStudy (anonymous):
I don't know how t do it
OpenStudy (xapproachesinfinity):
we have the explicit formula
\(\huge u_n=5^n-1\)
OpenStudy (xapproachesinfinity):
so\(\huge u_2=5^2-1\) yess?
OpenStudy (xapproachesinfinity):
just replacing n with 2
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OpenStudy (xapproachesinfinity):
so \(\huge u_2=24\)
OpenStudy (xapproachesinfinity):
let's see if yu can do \(\huge u_3\)
OpenStudy (anonymous):
Did you look at the attached file?
OpenStudy (xapproachesinfinity):
i don't really care about the attached file
OpenStudy (xapproachesinfinity):
that's just to check for the last thing to do
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OpenStudy (xapproachesinfinity):
what you need to focus on is finding the formula
from those terms
OpenStudy (anonymous):
I see what you're saying and i'm kind of getting it but it's not matching the answers
OpenStudy (xapproachesinfinity):
it doesn't matter for now okay
OpenStudy (xapproachesinfinity):
to find the general formula you have to derive by doing what are doing now
OpenStudy (anonymous):
okay so i need to figure out what u3 means right?
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OpenStudy (xapproachesinfinity):
yes!
OpenStudy (anonymous):
I'm so sorry but my session timed out and took me to a new question. I think i get it though.
OpenStudy (xapproachesinfinity):
by this point you should see the pattern
OpenStudy (xapproachesinfinity):
oh does not matter, what matters is you understand what you are doing
not the homework or quiz itself
OpenStudy (xapproachesinfinity):
oh hold i miss typed it...
correction in the next reply
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OpenStudy (xapproachesinfinity):
i just realize i was doing a wrong problem
it should be \(\huge u_n=5^{n-1}\) not \(\huge u_n=5^n-1\)you should correct me
on this from the beginning
OpenStudy (xapproachesinfinity):
i was dealing with something different
OpenStudy (xapproachesinfinity):
but we still have \(\huge u_2=5\\ \huge u_3=5^2 \\ \huge u_4=u^3\)
so apparently \(\huge u_3=5^2=5*5=5u_2\)
and \(\huge u_4=5*5^2=5u_3\) and you can check \(\huge u_5, u_6\) if you want
we generalize from this that \(\huge u_{n+1=5\huge{u_n}}\)
OpenStudy (xapproachesinfinity):
if you check you answer that you have you get the third one
OpenStudy (xapproachesinfinity):
the first one i mean D
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OpenStudy (xapproachesinfinity):
do you get it?
OpenStudy (xapproachesinfinity):
just look at the pattern
OpenStudy (xapproachesinfinity):
sorry about the confusion i went a wrong equation in the beginning
OpenStudy (xapproachesinfinity):
so the one you are looking for is
\(\huge u_{n+1}=5u_n\)
OpenStudy (xapproachesinfinity):
that's the recursive formula
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