POM TRI Triangles & Sequence RULE Question. Please help

Question

Jo constructs triangular patterns using dots.

Pattern 1 (has 3 dots forming a triangle)

Pattern 2 (has 6 dots forming a larger triangle)

Pattern 3 (has 10 dots forming an even larger triangle)

The pattern continues in the same geometric design.

How many dots are needed for the nth pattern?

Jo was born in 1953, she was wondering if she could make a triangular pattern out of exactly 1,953 dots. If she could, what would the pattern number be?

THANKS!

Answer 1

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

 

The described numbers are called the Triangular Numbers:

 

Triangle . . Dots

 Base . . . .Total

 . .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

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.

Answer 2

Pattern 1: 1+2 = 3 dots

Pattern 2: 1+2+3 = 6 dots

Pattern 3: 1+2+3+4 = 10 dots

...

Pattern n: 1+2+3+4+...n+(n+1) = ∑_k=1ⁿ⁺¹ (k) = (n+1)(n+2)/2 dots

 

To find n for 1953 dots, set

(n+1)(n+2)/2 = 1953, and simplify:

n² + 3n + 2 = 2(1953)

n² + 3n - 3904 = 0

(n + 64)(n - 61) = 0,

 

so that n=61 is the only positive solution. Check: 62*63/2 = 1953.