Quod Erat Demonstrandum

2011/08/24

無聊談談三角形數

Filed under: Fun,Junior Form Mathematics — johnmayhk @ 4:59 下午

●●

●●
●●●

●●
●●●
●●●●

$T_n = 1+2+3+\dots +n=\frac{n(n+1)}{2}$

$8T_n+1$ 一定是平方數（square number）。

$8T_n+1$
$=\frac{8n(n+1)}{2}+1$
$=4n^2+4n+1$
$=(2n+1)^2$

$T$ 是三角形數，則以下都是三角形數。

$9T+1$,
$25T+3$,
$49T+6$,
$81T+10$,
$121T+15$,
$169T+21$,

$9\times 10+1=91$ 也是三角形數，諸如此類。

$\frac{N(N+1)}{2}$

$\frac{(an+b)(an+b+1)}{2}$

$=\frac{a^2n^2+a(2b+1)n}{2}+\frac{b(b+1)}{2}$

$\frac{a^2n^2+a^2n}{2}+\frac{b(b+1)}{2}$

$=a^2(\frac{n(n+1)}{2})+\frac{b(b+1)}{2}$

$b=1$$a=3$，得 $(3)^2T+1=9T+1$ 是三角形數；

$b=2$$a=5$，得 $(5)^2T+3=25T+3$ 是三角形數；

$b=3$$a=7$，得 $(7)^2T+6=49T+6$ 是三角形數；

http://en.wikipedia.org/wiki/Triangular_number#Triangular_roots_and_tests_for_triangular_numbers

3 則迴響 »

1. You may find the following problem(s) interesting. Can a triangular number also a square number? If so, how many can you find? Can you find all of them?

迴響 由 koopakoo — 2011/08/25 @ 1:10 上午 | 回應

• Thank you Dr. Koopa! It is really an interesting problem and it is stunning to have

$(\frac{(1+\sqrt{2})^{2n}-(1-\sqrt{2})^{2n}}{4\sqrt{2}})^2$

for generating the ‘trianglar and square’ numbers!!

Please check the following for details:

http://www.cut-the-knot.org/do_you_know/triSquare.shtml

迴響 由 johnmayhk — 2011/08/25 @ 8:41 上午 | 回應

2. For every tri-square, they must satisfy the equation

n(n+1)/2=m^2

which can be simplified to Pell’s equation

(2n+1)^2=8m^2+1

then use the standard result of Pell’s equation.

迴響 由 Charlie — 2011/08/25 @ 12:11 下午 | 回應