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Triangular Numbers
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Triangular Numbers page.
© 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=
=n2+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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