# Thread: where does the n-1 go?

1. ## where does the n-1 go?

I I’m trying to understand that algebra in this problem. This is a Proof for checking solutions of recurrence.
I have a sequence defined as sn = 2n+4*3n n[IMG]file:///C:/Users/Owner/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]0
I have to prove that sn = 5sn-1 -6sn-2

Now the book does this as follows.
5(2n-1 + 4*3n-1) -6(2n-2+4*3 n-2) = (5*2-6) 2n-2 + (5*4*3 -6*4) 3 n-2 =Some other stuff I understand = 2n+4*3n
Where does the n-1 go, I’m not understanding the first part of this algebra?

2. Originally Posted by engineerairborne
I I’m trying to understand that algebra in this problem. This is a Proof for checking solutions of recurrence.
I have a sequence defined as sn = 2n+4*3n n[IMG]file:///C:/Users/Owner/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]0
I have to prove that sn = 5sn-1 -6sn-2

Now the book does this as follows.
5(2n-1 + 4*3n-1) -6(2n-2+4*3 n-2) = (5*2-6) 2n-2 + (5*4*3 -6*4) 3 n-2 =Some other stuff I understand = 2n+4*3n
Where does the n-1 go, I’m not understanding the first part of this algebra?

Either you write more carefully, with spaces and parentheses, or else learn the basics of LaTeX in this site and use it, otherwise it's almost impossible to understand what you wrote.

Tonio

3. You are missing some very important parentheses. Also what does "*" mean? Normally, it means "times" but if that is the case here, "4*3n" would be 12n and "s_n= 2n+ 3*4n" would be just 14n, probably not what you meant since it would have been much simpler to write. If "*" is supposed to indicate a power (standard is "^") is the power 3 or 3n? That is do you mean (3^4)n= 81n or 3^(4n)?

In any case, whether you have a function like f(x) or a sequence like s_n, you replace x or n in the formula by whatever replaces it in the expression.

That is, if f(x)= x^2 then f(x-1)= (x-1)^2. If s_n= 2n+ 3^(4n), then s_(n-1)= 2(n-1)+ 3^(4(n-1)= 2n- 2+ 3^(4n-4)= 2n- 2+ (3^(-4))(3^(4n))