Question

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

Answer 1

first time I've seen you here

Answer 2

\[a _{1}=1, a _{2}=2, a _{n+2}=4a _{n+1}-3a _{n}\]

Answer 3

Sorry, i take forever on this equation thingie.

Answer 4

ok, let's find \[a_3\] together

Answer 5

Okie dokie!

Answer 6

its like lil emu is in dsiguise helpin ya!;D lol i will leave ya alone now to get bak to hw

Answer 7

Okay. Got it.

Answer 8

\[a_3=a_{1+2}\]
which mean n=1
using equation

\[a_{2+1}=4a_{1+1}-3a_{1}\]

Answer 9

wait..so n does not equal 2?

Answer 10

no, I meant n=1

Answer 11

Okie.

Answer 12

\[a_{2+1}=4a_{1+1}−3a_1\]
\[=a_{3}=4a_{2}−3a_1\]

you know what a1 and a2 are

Answer 13

Oh coolio!!! We can plug those numbers in.

Answer 14

a_3=4(2)-3(1)=5

Answer 15

let see you find a_4

Answer 16

Okay, i got 5 for a_3

Answer 17

Would i have to make n equal to 2?

Answer 18

yes

Answer 19

I got 14 for a_4

Answer 20

Is that correct?

Answer 21

yes

Answer 22

YAAAYYY!!! now its soo simple for me!!

Answer 23

I got this, thank you thank you!!!!

Answer 24

done yet
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