Elementary question i feel.....

[berards]

years back i was playing pyramid solitare and suddenly i thought to myself.... how can i determine the amount circles in the formation of an equilateral triangle?

therefore for example:
**********************************0
*********************************0 0
********************************0 0 0
*******************************0 0 0 0 *next row is 5 then 6 and so on

If someone could tell me what the actual method of determining the amount of circles is without counting each one individually would be, i'd appreciate it.

i derived my own method and i am wondering how close i am to the logical equation.

 

[galactus]

It would appear you want the kth triangular number.

It is \(\displaystyle \frac{n(n+1)}{2}\)

Suppose, as in the triangle in your post, you want the 4th triangular number. this is the number of circles in a 4 tier 'pyramid'.

If n=4, then \(\displaystyle \frac{4(4+1)}{2}=10\). The number of circles in your trianglular 'pyramid'.

1,3,6,10,15,21,28,36,............................

If you want a pyramid 100 tiers high, you need the 100th triangular number. It would have 5050 circles.

 

[soroban]

Hello, berards!

Years back i was playing pyramid solitare and suddenly i thought to myself:
how can i determine the amount circles in the formation of an equilateral triangle?

For example:

. . \(\displaystyle \begin{array}{c}\circ \\ \circ\;\circ \\ \circ\;\circ\;\circ \\ \circ\;\circ\;\circ\;\circ \end{array}\)

. . next row is 5, then 6 and, so on

If someone could tell me what the actual method of determining the amount of circles is
without counting each one individually would be, i'd appreciate it.

i derived my own method and i am wondering how close i am to the logical equation.

\(\displaystyle \text{Here's one approach to deriving the formula.}\)


\(\displaystyle \text{You have }n = 4:\)

. . \(\displaystyle \begin{array}{c}\circ \\ \circ\;\circ \\ \circ\;\circ\;\circ \\ \circ\;\circ\;\circ\;\circ \end{array}\)


\(\displaystyle \text{Left-justify the circles:}\)

. . \(\displaystyle \circ\)
. . \(\displaystyle \circ\;\circ\)
. . \(\displaystyle \circ\;\circ\;\circ\)
. . \(\displaystyle \circ\;\circ\;\circ\;\circ\)


\(\displaystyle \text{Append an inverted copy of the triangle:}\)

. . \(\displaystyle \begin{array}{c} \circ\;\bullet\;\bullet\;\bullet\;\bullet \\ \circ\;\circ\;\bullet\;\bullet\;\bullet \\ \circ\;\circ\;\circ\;\bullet\;\bullet \\ \circ\;\circ\;\circ\;\circ\;\bullet \end{array}\)


\(\displaystyle \text{The rectangle has: }\:4 \times 5 \,=\,20\text{ circles.}\)

\(\displaystyle \text{The triangle has: }\:\frac{4\times5}{2} \,=\, 10\text{ circles.}\)


\(\displaystyle \text{Therefore, an equilateral triangle of side }n\text{ has: }\:\frac{n(n+1)}{2}\text{ circles.}\)


\(\displaystyle \text{The numbers: }\:1,3,6,10,15,21,28,36, 45, 55 \hdots \,\text{ are called }Triangular\;Numbers\)
. . \(\displaystyle \text{for obvious reasons.}\)

 

[berards]

ty both for help....


the way i came up with it was: n squared / 2 + n / 2