Question

.3: Finding Square Patterns

Define the sequence W so that W(n) is the number of white squares in Step n, and define
the sequence B so that B(n) is the number of black squares in Step n.

Write an equation for sequence W.
Write an equation for sequence B.
Are the sequences arithmetic, geometric, or neither? Explain how you know

Is the number of black squares ever larger than the number of white squares?
Explain how you know.

Answer 1

Explanation:

Step 1:

• B(1) = 1 * 2 = 2.

• W(1) = 3 * 4 - B(1) = 12 - 2 = 10.

Step 2:

• B(2) = 2 * 3 = 6.

• W(2) = 4 * 5 - B(2) = 20 - 6 = 14.

Step 3:

• B(3) = 3 * 4 = 12.

• W(3) = 5 * 6 - B(3) = 30 - 12 = 18.

Clearly, we have B(n) = n(n + 1) and W(n) = (n + 2)(n + 3) - B(n) = (n + 2)(n + 3) - n(n + 1).

Clearly, B(n) is neither A.P. nor G.P., as seen by the sequence 2, 6, 12, ...

However, W(n) is an A.P., as seen below:

W(n) = (n + 2)(n + 3) - n(n + 1)

= (n² + 5n + 6) - (n² + n)

= 4n + 6.

This is precisely an A.P. with common difference 4.

Set B(n) > W(n).

=> n(n + 1) > 4n + 6

=> n² + n > 4n + 6

=> n² - 3n - 6 > 0.

Clearly, since n² - 3n - 6 is a quadratic polynomial that concave upwards, there exists a n such that B(n) > W(n).