how do you work out the formula (nth term) for the triangular number sequence

Guest Jan 3, 2015

#2**+10 **

Thanks for that reference, Alan....

The "triangular" numbers are generated by just adding the sum of the first "N" digits

Here's another way of generating the "formula"

Ist, write down all the first N positive intergers.....and below that, write them in reverse order....so we have

1 2 3 4 5 ...... N-2 N-1 N

N N-1 N-2 N-3 N-4 .... 3 2 1 ...now...just add each column, this gives

N+1 N+1 N+1 N+1 N+1 .... N+1 N+1 N+1

Now the total sum is just (N+1) added N times = (N+1)(N)

But, we've got twice the sum we really need, so just divide this by 2 and we have

(N)(N+1) / 2

And, presto!!!....there's the "formula"

CPhill Jan 3, 2015

#2**+10 **

Best Answer

Thanks for that reference, Alan....

The "triangular" numbers are generated by just adding the sum of the first "N" digits

Here's another way of generating the "formula"

Ist, write down all the first N positive intergers.....and below that, write them in reverse order....so we have

1 2 3 4 5 ...... N-2 N-1 N

N N-1 N-2 N-3 N-4 .... 3 2 1 ...now...just add each column, this gives

N+1 N+1 N+1 N+1 N+1 .... N+1 N+1 N+1

Now the total sum is just (N+1) added N times = (N+1)(N)

But, we've got twice the sum we really need, so just divide this by 2 and we have

(N)(N+1) / 2

And, presto!!!....there's the "formula"

CPhill Jan 3, 2015