## Some Triangular number identities — Part 2

Prove the following relations, involving triangular numbers;

$T_{k + n} = T_{k} + T_{2 \, n \, m + n}$,     where     $k = 2 \, n \, m^2 + (2 \, n + 1) \,m$

T$_{n} = T_{n-1} + T_{m} + T_{k}$,     where     $n = (m^2 + k^2 + m + k)/2$

$T_{a+n} + T_{b+n} + T_{c+n} + T_{d+n} \; = \; T_{m - a} + T_{m - b} + T_{m - c} + T_{m - d}$

where     $m = n + (a+b+c+d)/2$

$T_{a} + T_{b} = ( \,T_{n^2 + n - 1} + T_{n^2 - n - 1} \,) \, ( \,T_{m^2 + m - 1} + T_{m^2 - m - 1} \,)$

where     $a = n^2 \, m^2 + n \, m - 1$,     $b = n^2 \, m^2 - n \, m - 1$

$T_{7 \, c + 1} + T_{c - 1} = (T_{7 \, n + 1} + T_{n - 1}) (T_{7 \, m + 1} + T_{m - 1})$

where     $c = 5 \,n \,m + n + m$

$T_{n} - T_{m} = 3^{2 \; \alpha} k$,

where     $n = 3^{2 \; \alpha} k + (3^{2 \; \alpha} - 1)/2$,     $m = 3^{2 \; \alpha} k - (3^{2 \; \alpha} + 1)/2$

$T_{n} + T_{m} = T_{m - 3^{2 \; \alpha} k} + T_{m + 3^{2 \; \alpha} k}$

where     $n = 3^{2 \; \alpha} k^2 + (3^{2 \, \alpha} - 1)/2$,     $m = 3^{2 \; \alpha} k^2 - (3^{2 \; \alpha} + 1)/2$