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Triangle numbers

$T_{n}=\frac{n(n+1)}{2}$

$T_{2n}:=3, 10, 21, 36, 55, 78, ...$

I noticed the following patterns

$2\cdot3=3!$

$2^2\cdot3\cdot10=5!$

$2^3\cdot3\cdot10\cdot21=7!$

And so on...

The general result for odd factorial

$n\ge1$

$$2^n\cdot{T_{2}}\cdot{T_{4}}\cdot{T_{6}}\cdots{T_{2n}}=(2n+1)!$$

How do we prove this claim result?


$T_{2n-1}:=1, 6, 15, 28, 45, 66, ...$

I noticed the following patterns

$2\cdot1=2!$

$2^2\cdot1\cdot6=4!$

$2^3\cdot1\cdot6\cdot15=6!$

And so on ...

The general result for even factorial

$n\ge1$

$$2^n\cdot{T_1}\cdot{T_3}\cdot{T_5}\cdots{T_{2n-1}}=(2n)!$$

How do we prove this claim result?

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  • 2
    $\begingroup$ Isn't it obvious from $T_n=\frac{1}{2}n(n+1)$ $\endgroup$ – almagest May 17 '16 at 18:14
  • $\begingroup$ Proof by induction. (Or just look at the factors of $(2n)!$ and note that $2T_{2k-1}=2k(2k-1)$.) $\endgroup$ – Semiclassical May 17 '16 at 18:19
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As you said, $T_n=\frac{n(n+1)}{2}$. Then substituting this in the product $P_{n}=T_{2}T_{4}T_{6}..T_{2n}$ gives $$P_n=\frac{2\cdot3}{2}\cdot\frac{4\cdot5}{2}\cdot\frac{6\cdot7}{2}\cdot\cdot\frac{2n\cdot(2n+1)}{2}=\frac{(2n+1)!}{2^n}$$ Similarly for the odd case. To make this more systematic, you could prove it by induction, showing that $\{P_n=\frac{(2n+1)!}{2^n}\}$ implies $\{P_{n+1}=\frac{(2n+3)!}{2^{n+1}}\}$. Since you know it to be true for $n=1$, it follows for all $n$.

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