Challenge #8 solution

…in which you were asked to find a formula for $\Delta_n$ , the nth 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!

$\begin{array}{ccccccccc} S & = & 1 & + & 2 & + & \cdots & + & n \\ +S & = & n & + & (n-1) & + & \cdots & + & 1 \\ \hline 2S & = & (n+1) & + & (n+1) & + & \cdots & + & (n+1) \end{array}$

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

$\begin{array}{rcl} 2S & = & \overbrace{(n+1) + (n+1) + \cdots + (n+1)}^{n} \\ 2S & = & n \cdot (n+1) \\ S & = & \frac{n \cdot (n+1)}{2}. \end{array}$

And we have our formula: $\Delta_n = \frac{n(n+1)}{2}$ . A few examples are in order to check that it works. When n = 3, $\Delta_3 = 3 + 2 + 1 = 6$ , and sure enough, $\frac{3 \cdot 4}{2} = 6$ . When n = 7, we have 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28, and $\frac{7 \cdot 8}{2} = 7 \cdot 4 = 28$ . Awesome!

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

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6 Responses to Challenge #8 solution

Andre says:

January 4, 2008 at 6:14 pm

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.
Brent says:

January 6, 2008 at 8:21 pm

Andre:

Right! And in general, this is part of a bigger pattern, namely that $\binom{n}{k} = \binom{n}{n-k}$ (that is, Pascal’s triangle is left-right symmetric). So $\binom{n+1}{n-1} = \binom{n+1}{n+1-(n-1)} = \binom{n+1}{2}$ .
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