Check out some similar questions!

endlesslove230 · Oct 23, 2011, 06:26 AM

$c_1=-1$ - the fist term
$c_{n 1}=3 c_n 1$

Unknown008 · Oct 24, 2011, 12:17 AM

I think there is a mistake in the problem you typed...

Is that:

$c_{n+1} = 3c_{n-1}$

?

endlesslove230 · Oct 24, 2011, 11:50 AM

Oh, sorry, I should have paid more attention

$c_(n+1) = 3 c_n + 1$

endlesslove230 · Oct 24, 2011, 11:51 AM

n+1 is meant to be subscript, but I don't know how to to write it.

ebaines · Oct 24, 2011, 12:25 PM

Your sequence goes like this:

-1, -2, -5, -14, -41, -122...

Here's an identity that can help you get started:

$<br /> \displaystyle\sum _{i=0} ^{n-1} x^i = \frac {1-x^n} {1-x}<br />$

For your problem set x = 3. This gives the sequence: 1, 4, 13, 40, 121, etc. Can you see how to modify this to get your sequence?

And to get subscripts in LaTex: enclose the item in squiggly brackets if it's more than one character. Same trick applies to superscripts too. Hence use C_{n+1} = 3C_n +1

to produce:

$<br /> C_{n+1} = 3C_n +1<br />$