Consider the sequence defined by b0 = 0, b1 = 1, b2 = 2, and bn = bn−1 + 2bn−2 − 2bn−3 for n ≥ 3. Determine the values of b3, b4, b5, b6, and a closed form expression for bn. (You will need to think about how to appropriately extend the type of closed form expression given above.)

To find the values of b3, b4, b5, b6, we can use the recursive formula bn = bn−1 + 2bn−2 − 2bn−3.

Let's start by finding b3:

b3 = b2 + 2b1 - 2b0
= 2 + 2(1) - 2(0)
= 2 + 2 - 0
= 4

Next, let's find b4:

b4 = b3 + 2b2 - 2b1
= 4 + 2(2) - 2(1)
= 4 + 4 - 2
= 6

Now, let's find b5:

b5 = b4 + 2b3 - 2b2
= 6 + 2(4) - 2(2)
= 6 + 8 - 4
= 10

Lastly, let's find b6:

b6 = b5 + 2b4 - 2b3
= 10 + 2(6) - 2(4)
= 10 + 12 - 8
= 14

The values of b3, b4, b5, b6 are 4, 6, 10, 14 respectively.

To find a closed form expression for bn, we can try to find a pattern in the sequence.

If we write out a few terms of the sequence, we have:

b0 = 0
b1 = 1
b2 = 2
b3 = 4
b4 = 6
b5 = 10
b6 = 14

It seems that the terms are increasing by multiples of 2: 1, 2, 4, 6, 10, 14. Let's examine the differences between the terms:

1 - 0 = 1
2 - 1 = 1
4 - 2 = 2
6 - 4 = 2
10 - 6 = 4
14 - 10 = 4

We can see that the differences are also increasing by multiples of 2: 1, 1, 2, 2, 4, 4. This suggests that the sequence can be represented by a quadratic function.

Let's try a quadratic function of the form bn = an^2 + bn-1, where a is a constant.

Substituting n=1 into the quadratic function, we have:

b1 = a(1)^2 + b0
1 = a(1) + 0
a = 1

So the quadratic function representing the sequence is:

bn = n^2 + bn-1

Now we can write the closed form expression for bn:

Answer: bn = n^2 + bn-1