Recent content by pavel329

1. I don't understand how to prove this. for all n≥1, 10n - 1 is divisible by 9. 3. I've done the basis step. Now I'm on the inductive step. I'm using (10k+1-1)/9=1. I don't know where to go from there. Using algebra just gets me down to 10k+1= 10. And I really don't think that's...

Thank you very much for your help.

I've only seen one example. sn = sn-1 + 6sn-2, s0 = 4 s1 = 7 However this equation put a 0 in place of the an and added the n≥2. Which maybe I shouldn't have paid attention to. But my final answer is: A=1 B=0. So an = 3n. Is this correct?

Ok so I now have it down to t=2,3. Now I assume i need to put that into an equation for the original problem. Which I have no clue how to do. But where does the n≥2 stand? Does that affect my final outcome?

I'm not sure how to factor anything with (n-1) or (n-2) as an exponent. The book turned a similar equation into t2-t-6. Which made no sense to me.

1. Solve the following recurrence relation. an - 5an-1 + 6an-2 = 0, n ≥ 2, a0 = 1, a1 = 3 3. My attempt I changed it to 0 = tn - 5tn-1 + 6tn-2 Don't know where to go from there.