# Generic triangular numbers sequence formula

I know that I can get the nth element of the following sequence

1   3   6  10  15  21


With the formula

(n * (n + 1)) / 2


where n is the nth number I want. How can I generalise the formula to get the nth element of the following sequences where by following sequences I mean

1 -> 1   3   6   10  15  21
2 -> 2   5   9   14  20
3 -> 4   8   13  19
4 -> 7   12  18
5 -> 11  17
6 -> 16


$$S_n=1+2+4+7+11+16+\cdots+t_n$$
$$S_n=\cdots\cdots\cdots1+2+4+7+11+16+\cdots+t_n$$
$$S_n-S_n=1+(2-1)+(4-2)+(7-4)+(11-7)+\text{ up to }n-1\text{ terms}-t_n$$
$$\implies t_n=1+\dfrac{(n-1)(1+n-1)}2$$
Now the $$r(1\le r\le6)$$th row has $$6+1-r$$ terms starting with $$1+\dfrac{r(r-1)}2$$