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Consider the sequence defined by  \(\begin{cases} s_0=0\\ s_1=3\\ s_n=6s_{n-1}-9s_{n-2} & \text{if }n\ge 2 \end{cases}\) 

Find a closed form for \(s_n\).

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Consider the sequence defines by \(\begin{cases} t_0=5\\ t_1=9\\ t_n=6t_{n-1}-9t_{n-2} & \text{if }n\ge 2 \end{cases} \)

Find the closed form for \(t_n\).

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I am struggling to try to solve these problems. Especially the second one. i listed the terms out, but i just cant seem to find the pattern. Help, hints, or the solution would be appreciated!

 
Guest Sep 16, 2018
 #1
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+1

1)-

 

 S(0) =0
S(1)=3
S(2)=18
S(3)=81
S(4)=324
S(5)=1215
S(6)=4374
S(7)=15309
0, 3, 18, 81, 324, 1215, 4374, 15309, 52488, 177147, 590490, 1948617, 6377292, 20726199, ...etc.
a_n = 3^(n - 1) (n - 1)  - The closed form.

 

2)-

 

t(0)=5
t(1)=9
t(2)=9
t(3)=-27
t(4)=-243
t(5)=-1215
t(6)=-5103
t(7)=-19683
5, 9, 9, -27, -243, -1215, -5103, -19683, -72171, -255879, -885735, -3011499....etc.
a_n = -3^(n - 1) (2 n - 7) - The closed form.

 
Guest Sep 16, 2018
 #2
avatar+26965 
+1

I get these to be:

 

sn = n3n   and tn = (5 - 2n)3n  where n starts from 0.  (i.e. n = 0, 1, 2, 3, ...)

 
Alan  Sep 16, 2018
 #3
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+1

1- Expand the following:
3^(n - 1) (n - 1) = n×3^n

Express 3^(n - 1) (n - 1) as a difference of fractions.
3^(n - 1) (n - 1) = 3^(n - 1) n - 3^(n - 1):
3^(n - 1) n - 3^(n - 1) = n*3^n

 

2- Expand the following:
-3^(n - 1) (2 n - 7) = 3^n (5 - 2 n)

Express -3^(n - 1) (2 n - 7) as a difference of fractions.
-3^(n - 1) (2 n - 7) = -(3^(n - 1) (-7)) - 3^(n - 1)×2 n:
-(-7×3^(n - 1)) - 2×3^(n - 1) n = 3^n (5 - 2 n)

Multiply -1 and -7 together.
×3^(n - 1) - 2×3^(n - 1) n = 3^n (5 - 2 n)

Distribute 3^n over 5 - 2 n.
3^n (5 - 2 n) = 3^n×5 + 3^n (-2 n):

7×3^(n - 1) - 2×3^(n - 1) n = 5×3^n - 2×3^n n =(5 - 2n)*3^n

 
Guest Sep 16, 2018
edited by Guest  Sep 16, 2018
 #4
avatar+26965 
+1

Hmmm!   

 

n        3^(n-1)(n-1)                n3^n

0        3^(-1)(-1) = -1/3         0*3^0 = 0

1        3^(1-1)(1-1) = 0         1*3^1 = 3

2        3^(2-1)(2-1) = 3         2*3^2 = 18

...etc.

 
Alan  Sep 17, 2018

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