Match the Product

Number Theory Level 4

$\large P = \prod_{k=0}^{m-1}\left[m(m+1)-2T_k\right]$

where $T_k$ is the $k$ th triangular number and $m=\left \lceil \dfrac n2 \right \rceil$

Then $P$ equals which of the following:

$\displaystyle 2\prod_{x=0}^{m-1}(m^2-x^2)$ for odd $n$
$2n!$
$\displaystyle 2\prod_{x=0}^{m-1}(m^2-x^2)$ for even $n$
$n!$

Notations:

$T_k=\dfrac {k(k+1)}2$ , where $k$ is a non-negative integer, denotes the triangular numbers 0,1,3,6,10,15....
$\lceil \cdot \rceil$ denotes the ceiling function .

Both 1 and 3 Only 2 Both 2 and 3 Only 3 Both 2 and 4 Only 1 Only 4

1 solution

Mark Hennings
Jul 12, 2017

We have $\begin{aligned} P & = \prod_{k=0}^{m-1} \big(m(m+1) - 2T_k\big) \; = \; \prod_{k=0}^{m-1}\big(m(m+1) - k(k+1)\big) \\ & = \prod_{k=0}^{m-1}(m-k)(m+k+1) \; = \; \prod_{k=0}^{m-1}(m-k) \prod_{k=0}^{m-1}(m+k+1) \\ & = m! \times \frac{(2m)!}{m!} \; = \; (2m)! \; = \; \left\{ \begin{array}{ll} n! & n \mbox{ even} \\ (n+1)! & n \mbox{ odd} \end{array} \right. \end{aligned}$

Thus options $(2)$ and $(4)$ are not always true. On the other hand $\begin{aligned} Q & = 2 \prod_{x=0}^{m-1} (m^2 - x^2) \; = \; 2\prod_{x=0}^{m-1} (m-x)(m+x) \\ & = 2 \times m! \times \frac{(2m-1)!}{(m-1)!} \; = \; 2m\times (2m-1)! \; = \; (2m)! \end{aligned}$ for all $n \ge 1$ .

The answer should be $\boxed{(1) \mbox{ and } (3)}$ .

Thank you for an excellent solution

Mrigank Shekhar Pathak - 3 years, 11 months ago

Match the Product

1 solution

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