Rewrite the determinant of each triangular number in terms of combination (choose any eg 3,6,10,15) and, **hence**, establish a determinant for the nth triangular number with elements **in terms of n only**. Show, by proof that this determinant always produces a triangular number.

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The elements of Pascal's triangle can be written as:

`_0C_0`

`_1C_0` ` ``_1C_1`

`_2C_0` `_2C_1` `_2C_2`

`_3C_0` `_3C_1` `_3C_2` `_3C_3`

etc...

The triangular numbers are found along the diagonal beginning at `_2C_0` through `_3C_1` ,`_4C_2` ,`_5C_3` ,etc...

**A rule for the triangular numbers using combinations is** `_nC_(n-2)`

An algebraic rule for the triangular numbers is `(n(n-1))/2` : this generates all triangular numbers for `n>=1,n in NN` .

We show that `_nC_(n-2)=(n(n-1))/2` :

`_nC_(n-2)=(n!)/((n-2)!(n-(n-2))!)`

`=(n(n-1)(n-2)!)/((n-2)!2!)`

`=(n(n-1))/2` as required.

**Sources:**

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The formula `_nC_(n-2)` gives the triangular number `T_(n-1)` ; (thus the first triangular number `T_1=_2C_0=1` )

We can also write the formula as follows:

`T_n=`` _(n+1)C_(n-1)` if you prefer.

Proof: `T_n=(n(n+1))/2=((n+1)!)/((n-1)!(n+1-(n-1))!)`

`=` `_(n+1)C_(n-1)`

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