Question

Compute the first five terms of the sequence defined recursively by a1=1, a2=3; ak=(k-1)/(k-2). k≥3. The first term is 1, the second is 3.

Answer 1

this is actually not a recursion the way it is written. it says
\[a_k=\frac{k-1}{k-2}\] unless there is something missing

Answer 2

Oh I see. It's actually:
ak=(ak−1)/(ak−2)

Answer 3

ooh ok

Answer 4

\[_k=\frac{a_{k-1}}{a_{k-2}}\]

Answer 5

yes!

Answer 6

\[a_1=1,a_2=3\] then
\[a_3=\frac{a_2}{a_1}=\frac{3}{1}=3\]

Answer 7

and
\[a_4=\frac{a_3}{a_2}=\frac{3}{3}=1\]

Answer 8

and finally
\[a_5=\frac{a_4}{a_3}=\frac{1}{3}=\frac{1}{3}\]

Answer 9

Oh I see what you did. That makes sense :) Thank you so much! Would you mind helping me with another?

Compute the first five terms of the sequence defined recursively by A1=1; An=((2An-1)+2)/(-An-1). n≥2.

Answer 10

let me try to write it

Answer 11

\[An=\frac{2A_{n-1}+2}{-A_{n-1}}\]

Answer 12

like that?

Answer 13

first replace A by 1 to see what the next term will be. it will be
\[A-2=\frac{2\times 1+2}{-1}=-4\]

Answer 14

i meant
\[A_2=-4\]

Answer 15

Exactly like that!

Answer 16

the replace A by -4 to find
\[A_3=\frac{2\times -4+2}{-(-4)}=-\frac{3}{2}\]

Answer 17

and so on

Answer 18

Perfect I see exactly what you're doing. Thank you so much. :) You're great!!