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Find the first 6 terms of each sequence: I will write down in equation editor.

Mathematics
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first time I've seen you here
\[a _{1}=1, a _{2}=2, a _{n+2}=4a _{n+1}-3a _{n}\]
Sorry, i take forever on this equation thingie.

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Other answers:

ok, let's find \[a_3\] together
Okie dokie!
its like lil emu is in dsiguise helpin ya!;D lol i will leave ya alone now to get bak to hw
Okay. Got it.
\[a_3=a_{1+2}\] which mean n=1 using equation \[a_{2+1}=4a_{1+1}-3a_{1}\]
wait..so n does not equal 2?
no, I meant n=1
Okie.
\[a_{2+1}=4a_{1+1}−3a_1\] \[=a_{3}=4a_{2}−3a_1\] you know what a1 and a2 are
Oh coolio!!! We can plug those numbers in.
a_3=4(2)-3(1)=5
let see you find a_4
Okay, i got 5 for a_3
Would i have to make n equal to 2?
yes
I got 14 for a_4
Is that correct?
yes
YAAAYYY!!! now its soo simple for me!!
I got this, thank you thank you!!!!
done yet ??????????????

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