Solving: a_n=3a_n-1 + 2n to cn + d = 3(c(n-1) + d) + 2n

[yakin]

Did not understand how this \(\displaystyle a_n=3a_n-1 + 2n\) changed into \(\displaystyle cn + d = 3(c(n-1) + d) + 2n\)

PS: By the way, i used MathType syntax it didn't change anything?

 

[MarkFL]

You need to wrap your code within the math tags. The simplest way is to click the $\Sigma$ button on the toolbar (above the post field where you compose your post) to generate the tags, and then type your code in between them.

They are letting:

\(\displaystyle a_n=p_n=cn+d\)

And so wherever $a_n$ occurs in the recurrence, you replace it with $cn+d$. Do you see this is what they have done?

Now it is just a matter of using the method of undetermined coefficients to find $c$ and $d$. :D

 

[yakin]

Ok i got that, then they have simplified that expression to this expression how did they get this
\(\displaystyle (2 + 2c)n + ( 2d – 3c) = 0\)

 

[MarkFL]

Ok i got that, then they have simplified that expression to this expression how did they get this
\(\displaystyle (2 + 2c)n + ( 2d – 3c) = 0\)

They distributed, and then collected like terms and then factored. Give it a try, and if you can't get there, I will post my work. :D

 

[yakin]

I got it sir. Thanks!

 

[MarkFL]

I got it sir. Thanks!

We're pretty laid back and informal here...you can just call me Mark if you like. (Handshake)

 

[yakin]

We're pretty laid back and informal here...you can just call me Mark if you like. (Handshake)

Ok Mark.