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Triangular Numbers
Note: This is the alternative Triangular Numbers page, for WWW browser
which cannot display special symbols. In particular, I am using ^2 for
"squared" in this version. Please let me know if this is working for you or
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© Copyright 1999, Jim Loy
A triangular number is the
number of dots (or other objects) that can be arranged in a triangle as in the
diagram on the left. This is how bowling pins, pool balls, or snooker balls are
arranged. The diagram shows the first six triangular numbers, 1,3,6,10,15,21.
Notice that they are also the numbers 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5, ... In
other words, each triangular number is a particularly simple arithmetic series
(progression). See How To Be A Little Gauss. Also
notice that the sequence is odd, odd, even, even, odd, odd. It is easy to show
that that pattern goes on forever.
If you mess around with
triangular numbers, for a while, you may notice that the sum of any two
consecutive triangular numbers seems to be a square. 1+3=4, 3+6=9, 6+10=16,
etc. Do you think that is is always true? On the right, I have rearranged the
dots in two consecutive triangular numbers. Forget for a moment that these are
the 5th and 6th triangular numbers, as this works for any two consecutive
triangular numbers. This arrangement of dots is indeed a square, and is for any
pair of consecutive triangular numbers. In fact the nth triangular number plus
the (n+1)th triangular number is (n+1)^2.
That was easy to prove geometrically. Can we prove it algebraically?
Well, we have an equation for an arithmetic series. One form of this is
S=na+nd(n-1)/2, where S is the sum, n is the number of terms, a is the first
term, and d is the difference between terms. The first term and the difference
are always 1 here. So, S=n+n(n-1)/2 or S=n(n+1)/2. The sum of the nth
triangular number and the (n+1)th triangular number is:
n(n+1)/2 + (n+1)(n+2)/2=
=n^2+2n+1
=(n+1)^2
So, we have proved it two ways.
P.S. I know that (n+1)th should be (n+1)st.
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