…in which you were asked to find a formula for , the *n*th triangular number.

We can use the same technique as before, but this time some variables will be involved instead of just numbers. That’s OK, variables don’t scare us!

So we know that twice the sum *S* is the same as adding together *n* copies of (*n* + 1).

And we have our formula: . A few examples are in order to check that it works. When *n* = 3, , and sure enough, . When *n* = 7, we have 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28, and . Awesome!

39.953605
-75.213937

### Like this:

Like Loading...

*Related*

##
About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

Hi.

I already knew this way of finding out a formula for the nth triangular number but I looked for another way of getting it. I know this post was published almost a year ago but I loose nothing telling you other way to get to n(n+1)/2.

I found the triangular number series in the pascal triangle. We can see it here: http://haacked.com/images/PascalTriangle.gif

Looking to that sequence and comparing the order of each term I found that the nth triangular number is equal to n+1 C n-1 (I’m sorry I don’t know how to write it properly here), but that is the same as (n+1)!/(n+1-n+1)!*(n-1)! and if we simplify that:

(n+1)! (n+1)(n)(n-1)! n(n+1)

_______________ = _______________ = _________

(n+1-n+1)!*(n-1)! 2*(n-1)! 2

which is the same.

It’s just another way of finding a formula for delta n.

I’m sorry about my poor english.

Andre:

Right! And in general, this is part of a bigger pattern, namely that (that is, Pascal’s triangle is left-right symmetric). So .

Pingback: Math Teachers at Play #21 « The Math Less Traveled

Pingback: Chessboard counting: solutions and further challenges « The Math Less Traveled

Pingback: Triangunit divisors and quadratic reciprocity | The Math Less Traveled

Pingback: Triangular number equations via pictures | The Math Less Traveled