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Mathematics 10 Online
OpenStudy (anonymous):

Given the explicit formula u(n) = 5^n-1, where u(1) = 1, find the recursive formula for u(n).

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

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

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

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?

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

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

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

OpenStudy (xapproachesinfinity):

any further help?

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