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find the first 4 terms
\[a_1=\frac{3}{2}; a_{n+1}=\frac{n^2+1}{n(a_n)}\]
 one year ago
 one year ago
find the first 4 terms \[a_1=\frac{3}{2}; a_{n+1}=\frac{n^2+1}{n(a_n)}\]
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.1
i got my answer but apparently i was wrong this from @lopez_hatesmath
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
maybe some fresh eyes would help
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
nvm sorry to bother you
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
it was this \[a_1=\frac{3}{2}; a_{n+1}=\frac{n^2+1}{n}\times a_n\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
that is, the \(a_n\) was in the NUMERATOR
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
so now it is not so bad replace \(n=1\) on the right hand side to get \(a_2\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
you get \[a_2=\frac{1^2+1}{1}\times \frac{3}{2}\] \[a_2=2\times\frac{3}{2}\] \[a_2=3\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
now replace \(n\) by 2 on the right hand side to get \[a_3=\frac{2^2+1}{2}\times 3\] \[a_3=\frac{5}{2}\times 3\] \[a_3=\frac{15}{2}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
how are we doing so far?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
one more? \[a_4=\frac{3^2+1}{3}\times \frac{15}{2}\] \[a_4=\frac{10}{3}\times \frac{15}{2}\] \[a_4=25\]
 one year ago

lopez_hatesmathBest ResponseYou've already chosen the best response.0
okayy i got it!
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
not so bad that is the idea, i thought the term was in the denominator which is why i was screwing it up
 one year ago

lopez_hatesmathBest ResponseYou've already chosen the best response.0
gotcha thanks for the help man.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yw good luck with the next one, but it works the same so you should be good to go
 one year ago
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