Can somene tell if I have started this right please.

[ptex]

I need to find the explicit formuala for this recursive sequence.

a[sub]1[/sub] = 3
a[sub]2[/sub] = 7
a[sub]n[/sub] = 7a[sub]n-1[/sub] - 10a[sub]n-2[/sub]
a[sub]n[/sub] - 7a[sub]n-1[/sub] + 10a[sub]n-2[/sub] = 0
t[sup]n[/sup] - 7k[sup]n-1[/sup] + 10t[sup]n-2[/sup] = 0
t[sup]2[/sup] - 7t + 10 = 0
(t - 2)(t - 5)
(t = 2)(t = 5) :confused: 
if that is right I can move on.

 

[gnome]

Move on.

 

[ptex]

Thank you I will go ahead and stop if I have more questions I will stop. Ok?

 

[ptex]

Is this the formula?

a[sub]n[/sub] = 8/5(2)[sup]n[/sup] - 1/15(5)[sup]n[/sup]

 

[gnome]

No. Probably just a mistake someplace in your algebra.

You know, you can check your solution by using the recurrence to determine the value of a₃ and comparing that to the value generated by your formula.

 

[ptex]

To find B;

2(4A + 25B = 7) = 8A + 50B = 14
4(2A + 5B =  3) = 8A + 20B = 12
30B = 2
B = 1/15

To find A;

2A + 5B = 3
2A + 1/5B = 3
10A + 1 = 15
10A = 16
A = 16/10
A = 8/5

A = 8/5 B= 1/15
?

 

[gnome]

> 2A + 5B = 3
> 2A + 1/5B = 3 ...what is this? How did 5B become 1/5 B? You already know B. Plug it into the first eqn & solve for A.

 

[ptex]

B = 15 and A = 16?

 

[gnome]

No.

This was correct:

2(4A + 25B = 7) = 8A + 50B = 14
4(2A + 5B = 3) = 8A + 20B = 12
30B = 2
B = 1/15

Now use this equation to find A :

2A + 5B = 3

 

[ptex]

I think

a[sub]3[/sub] = 19

 

[gnome]

You think?

You're supposed to know.

 

[ptex]

Is A=1/2 if so

the formula would be a[sub]n[/sub] = 1/2(2[sup]n[/sup]) - 1/15(5[sup]n[/sup])

 

[gnome]

I don't understand what is confusing you.

You have already solved the simultaneous equations:

2(4A + 25B = 7) = 8A + 50B = 14
4(2A + 5B = 3) = 8A + 20B = 12
30B = 2
B = 1/15

so you know that B=1/15.

So now take one of your original equations (either one) and solve for A. Show me what you're doing, step by step.

 

[ptex]

woops

2A + 5B = 3
2A + 5(1/15) = 3
2A + 1 /3 = 3
2A = 3 1/3
A = 3 1/3 / 2
A = 5/3

 

[gnome]

[tex]B =\frac{1}{15}[/tex]

 

[ptex]

2a + 5b = 3
2a + 5(1/15) = 3
2a + 1 /3 = 3
2a = 3 1/3
A = 3 1/3 / 2
A = 5/3

 

[gnome]

2a + 5b = 3
2a + 5(1/15) = 3
2a + 1 /3 = 3
2a = 3 1/3 <<<<check this again
A = 3 1/3 / 2
A = 5/3

 

[ptex]

2a + 5b = 3
2a + 5(1/15) = 3
2a + 1 /3 = 3
2a = 10/3
A = 10/3 / 2
A = 5/3

 

[gnome]

2a + 5b = 3
2a + 5(1/15) = 3
2a + 1 /3 = 3
...- 1/3 = -1/3
2a = 8/3
A = 4/3

 

[ptex]

Thank you
the 1/3 turns negitive? then ?

 

[gnome]

The 1/3 doesn't turn negative. You subtract 1/3 from both sides of the equation. Then you divide both sides of the equation by 2. Then the equation is solved. A = 4/3.

So, you've never studied algebra and yet you are interested in solving linear homogeneous recurrence relations with constant coefficients. Very interesting.

 

[ptex]

The last time I took algebra was 10 years ago. This class Discrete Mathematics is a requirerment for my degree (all I want to do is write code). Yeah I know its a long time to get a degree but I have a bit of a skiing career so I can't go to school during the winter. Anyway thank you very much now I must move on to the second problem and try to recall algebra which by the seems more defficult then some of this stuff.