The tree sides of a triangle are of lengths $l$, $m$, and $n$, respectively. The numbers $l, m,$ and $n$ are positive integers.
$l \le m \le n$
Find the number of different triangles of the described kind for a given 'n'. ( Take $n = 1, 2, 3, 4, 5, ..., n)$
Find a general law governing the dependence of the number of triangles on $n$.
from Mathematics and Plausible Reasoning .
what i find is :
if i write the three sequence
1- n . (n)
2-number of all possibilities of l and m for each n .(P)
3- number of possible triangle that we can construct according to triangle iniquity. (T)
is
$p = (1,2,6,10,15,21,28,26,45,55,...)$
$T= (1,2,4,6,9,12,16,20,25,30,36,42,...)$
$n= (1,2,3,4,5,6,7,8,9,10,11)$
some observation about these sequence
1- for $P$ :
* it's increasing by n
*$P_n$ - $T_n-1$ = $T_n$
such that $P_n$ = (number of possible values for l and m for such n )
$T_n$ = number of possible triangles for n
2- For $T$
$1,1+1,2+2,2+2+2,2+2+2+3,2+2+2+3+3,2+2+2+3+3+4,2+2+2+3+3+4+4,2+2+2+3+3+4+4+5,2+2+2+3+3+4+4+5+5,...$
How can i derive general law ??
By the triangle inequality, for $l,m,n$ to make a triangle, $l+m>n$.
Fix $n$ and $m$, then we know that $n+1-m\leq l\leq m$. If $n+1-m\leq m$, then $\frac{n+1}{2}\leq m$ and these restrictions are not empty.
In such a case, there are $m-(n+1-m)+1=2m-n$ possible values for $l$. Now, we have two cases:
If $n$ is odd, then $\frac{n+1}{2}$ is a natural number, so we need: \begin{align*} \sum_{m=\frac{n+1}{2}}^n(2m-n)&=2\left(\frac{n(n+1)}{2}\right)-2\left(\frac{\frac{n-1}{2}\cdot\frac{n+1}{2}}{2}\right)-n\left(n-\frac{n-1}{2}\right)\\ &=n(n+1)-\frac{1}{4}(n^2-1)-\frac{1}{2}n(n+1)\\ &=(n+1)\left(\frac{1}{2}n-\frac{1}{4}(n-1)\right)\\ &=\frac{1}{4}(n+1)^2 \end{align*}
If $n$ is odd, then $\frac{n+1}{2}$ is not a natural number, so $\frac{n+1}{2}\leq m$ reduces to $\frac{n}{2}+1\leq m$. I'll leave the rest of the derivation as an exercise at this point.