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.
The elements of Pascal's triangle can be written as:
`_1C_0` ` ``_1C_1`
`_2C_0` `_2C_1` `_2C_2`
`_3C_0` `_3C_1` `_3C_2` `_3C_3`
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` :
`=(n(n-1))/2` as required.
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.