Steve S. answered • 01/29/14

Tutor

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Tutoring in Precalculus, Trig, and Differential Calculus

Let's find the sum of an arithmetic series (we'll need it later):

S = m + (m-1) + (m-2) + ... + . 3 . . . + . 2 . + .1

S = 1 . + . 2 . + . . 3 . . + ... + (m-2) + (m-1) + m

2S=m(m+1)

S = m(m+1)/2

S = m + (m-1) + (m-2) + ... + . 3 . . . + . 2 . + .1

S = 1 . + . 2 . + . . 3 . . + ... + (m-2) + (m-1) + m

2S=m(m+1)

S = m(m+1)/2

The described numbers are called the Triangular Numbers:

Triangle . . Dots

**B**ase . . . .

**T**otal

. .1 . . . . . . 1

. .2 . . . . . . 2 + 1 = 3

. .3 . . . . . . 3 + 2 + 1 = 6

. .4 . . . . . . 4 + 3 + 2 + 1 = 10

. .5 . . . . . . 5 + 4 + 3 + 2 + 1 = 15

...

. .B . . . . . . B(B+1)/2

In the problem description they assign an index of 1 to B = 2 and they want to know how many dots are in the n-th pattern.

B = n + 1

T(n) = (n+1)(n+2)/2 = number of dots in the n-th pattern.

1953 = (n+1)(n+2)/2

3906 = n^2 + 3n + 2

3906 = n^2 + 3n + 2

0 = n^2 + 3n - 3904 = (n + 64)(n - 61)

Negative n makes no sense, so answer is pattern n = 61.

This is a triangle with 62 dots in the base and 62(62+1)/2 = 1953 total dots.

BTW, the 3rd diagonals in Pascal's Triangle are the Triangular Numbers.