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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6 Responses to Challenge #8 solution

  1. Andre says:

    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.

  2. Brent says:

    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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