$\begin{array}{1 1} ^{2n+1}C_3 \\ n^3+n^2+2 \\ ^{2n+1}C_3-2.^nC_3+1 \\ n^3 \end{array}$

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To form a $\Delta$ **Case 1:**

we take any 2 points out of n+1 points $A_1,A_2,......A_n, and \:O$ and

one point out of the $n$ points $B_1,B_2,........B_n$

or

we take any 2 points out of n+1 points $B_1,B_2,......B_n$ and $O$ and

one point out of the $n$ points $A_1,A_2,........A_n$

$\therefore$ These $\Delta^s$ are $ ^{n+1}C_2\times n$ in number each case.

This includes those triangles,which are formed with one vertex at O and one

point on each line.These triangles are $n\times n=n^2$ in number.

i.e. These $n^2\:\Delta^s$ are counted twice. once with case 1 and once again with case 2.

$\therefore$ The required no. of $\Delta^s$ = $ ^{n+1}C_2 \times n+^{n+1}C_2\times n\:-\:n^2$

$=2.n.^{n+1}C_2-n^2$

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

$=n^3+n^2-n^2=n^3$

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