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**Mark35****Member**- Registered: 2016-12-26
- Posts: 2

1. Let

for alla. Write down the values of a_{1}, a_{2} and a_{3}.

b. Write down the values of A(1), A(2) and A(3) defined by the recurrence relation: A(0) = -1, A(k) = 3A(k - 1) - 2k + 7, k \geq 1

c. Show that A(k) = a_{k} is a solution of the recurrence relation for all values of k \geq 1.

2. Write down all derangements of the set \left\{ a,b,c,d \right\} and show that the number of derangements is the same as predicted by the recurrence D(n) = (n - 1)(D(n - 2) + D(n - 1)) with initial values D(1) = 0 and D(2) = 1.

3. Solve the recurrence relation A(n) = 6A(n - 1) - 11A(n - 2) + 6A(n - 3) subject to initial values A(1) = 2, A(2) = 6, A(3) = 20.

[Latex fixing by bobbym]

*Last edited by Mark35 (2016-12-26 23:34:06)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Latex needs to be put between the math tags for it to work here.

Over here it is the rule that you should show your work then people know what you need help on.

I will point you in the right direction.

1) This is just a matter of substituting numbers for the variable k.

For a1 you would put a 1 in every k and evaluate.

Can you evaluate that now?

2) Here are the derangements of {a,b,c,d}.

{b, a, d, c}

{b, c, d, a}

{b, d, a, c}

{c, a, d, b}

{c, d, a, b}

{c, d, b, a}

{d, a, b, c}

{d, c, a, b}

{d, c, b, a}

you can now answer the question.

3) Have you been taught a method to solve recurrences like that yet?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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