## pattern of black and white triangles: prove relationship

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.
little_dragon
Posts: 226
Joined: Mon Dec 08, 2008 5:18 pm
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### pattern of black and white triangles: prove relationship

There's a picture that looks kind of like this:
picture:
/\
Row 1     /  \
/\\\/\
Row 2   /  \/  \
/\\\/\\\/\
Row 3 /  \/  \/  \
The picture in my book has more lines, but they all have the same pattern of alternating black and white triangles, always with white ones on the ends of the rows.

a) Find a formula for each of the following:
i) the number of triangles in the nth row,
ii) the total number of triangles in the first n rows,
iii) the total number of white triangles in the first n rows, and
iv) the total number of black triangles in the first n rows.
b) Show algebraically that your answer to (a)(ii) is the sum of your answers to (a)(iii) and (iv).

a)i) The counting pattern is n = 1, triangles in that row r = 1; n = 2, r = 3; n = 3, r = 5; n = 4, r = 7; n = 5, r = 9; n = 6, r = 11; n = 7, r = 13; and n = 8, r = 15. So you always get two more in the next line. If I made the one side sort of "spiky" by adding another black triangle, each row would have that row's number of diamonds, or twice that row's number of triangles. If I take that extra triangle away, then I get r = 2n - 1.

a)ii) The sums are n = 1, total number of triangles t = 1; n = 2, t = 1 + 3 = 4; n = 3, t = 4 + 5 = 9; n = 4, t = 9 + 7 = 16; n = 5, t = 16 + 9 = 25; n = 6, t = 25 + 11 = 36; n = 7, t = 36 + 13 = 49; and n = 8, t = 49 + 15 = 64. This is just t = n2, right?

a)iii) The sums are n = 1, white triangles w = 1; n = 2, w = 1 + 2 = 3; n = 3, w = 3 + 3 = 6; n = 4, w = 6 + 4 = 10; n = 5, w = 10 + 5 = 15; n = 6, w = 15 + 6 = 21; n = 7, w = 21 + 7 = 28; and n = 8, w = 28 + 8 = 36. For each row, I'm adding that row's number to the total, and I learned before that the total of the first n numbers is (1/2)(n)(n + 1). So can I use that here?

a)iv) There's always one less black triangle for each row, so I took the total of the white triangles for n rows and subtracted one for each of the n rows: (1/2)(n)(n + 1) - n.

b) By "algebraically", do they just mean to add my formulas and make sure they match?

Thanks!

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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Your work looks good so far, but you might be expected to simplify the expression for your last rule.

. . . . .$\frac{1}{2}(n^2\, +\, n)\, -\, n$

. . . . .$\frac{n^2\,+\,n}{2}\, -\, \frac{2n}{2}$

. . . . .$\frac{n^2\, -\, n}{2}$
b) By "algebraically", do they just mean to add my formulas and make sure they match?
I believe so. Add your expressions for (a-iii) and (a-iv), and make sure you can simplify or rearrange the result to match your expression for (a-ii):

. . . . .$\left(\frac{1}{2}(n^2\, +\, n)\right)\, +\, \left(\frac{n^2\, -\, n}{2}\right)$

. . . . .$\frac{(n^2\, +\, n)\, +\, (n^2\, -\, n)}{2}$

. . . . .$\frac{2n^2}{2}$

...and so forth.

little_dragon
Posts: 226
Joined: Mon Dec 08, 2008 5:18 pm
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Thanks!