Results 1 to 3 of 3

- February 21st 2010, 01:01 PM #1

- Joined
- Feb 2010
- Posts
- 1

## where does the n-1 go?

I I知 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知 not understanding the first part of this algebra?

- February 21st 2010, 06:46 PM #2

- Joined
- Oct 2009
- Posts
- 4,261
- Thanks
- 2

- February 22nd 2010, 03:30 AM #3

- Joined
- Apr 2005
- Posts
- 16,976
- Thanks
- 2013

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))