1

$$a_n = R_1a_{n-2} + R_2a_{n-3} + R_3a_{n-4} + CD^{n-4} \quad\text{ for } n\ge 4$$

I'm a little confused as to whether move the function around so that i solve the left hand side first for the equation below

$$a_n - (R_1 a_{n-2} + R_2 a_{n-3} + R_3 a_{n-4}) = CD^{n-4}$$

or treat the constant like the other three terms

DanielV
  • 23,556
PlayDis
  • 13

1 Answers1

0

Writing

$$a_n - (R_1 a_{n-2} + R_2 a_{n-3} + R_3 a_{n-4}) = C~D^{n-4}$$

Is actually a very good idea. You can also write

$$a_{n+1} - (R_1 a_{n-1} + R_2 a_{n-2} + R_3 a_{n-3}) = CD^{n-3} = DC~D^{n-4}$$

Can you see how to cancel out the $D^{n-4}$ now?

DanielV
  • 23,556