Editing Arithmetic progression (section)

==Product==
The [[product (mathematics)|product]] of the members of a finite arithmetic progression with an initial element ''a''<sub>1</sub>, common differences ''d'', and ''n'' elements in total is determined in a closed expression

:<math>\begin{align}
a_1 a_2 a_3 \cdots a_n &= a_1 (a_1+d) (a_1+2d) \cdots (a_1+(n-1)d) \\[1ex]
&= \prod_{k=0}^{n-1} (a_1+kd)
= d^n \frac{\Gamma{\left(\frac{a_1}{d} + n\right)}}{\Gamma{\left( \frac{a_1}{d} \right)}}
\end{align}</math>

where <math>\Gamma</math> denotes the [[Gamma function]]. The formula is not valid when <math>a_1/d</math> is negative or zero.

This is a generalization of the facts that the product of the progression <math>1 \times 2 \times \cdots \times n</math> is given by the [[factorial]] <math>n!</math> and that the product

:<math>m \times (m+1) \times (m+2) \times \cdots \times (n-2) \times (n-1) \times n </math>

for [[Natural number|positive integer]]s <math>m</math> and <math>n</math> is given by

:<math>\frac{n!}{(m-1)!}.</math>

===Derivation===

:<math>\begin{align}
a_1a_2a_3\cdots a_n &=\prod_{k=0}^{n-1} (a_1+kd) \\[2pt]
&= \prod_{k=0}^{n-1} d\left(\frac{a_1}{d}+k\right) \\[2pt]
&= d \left (\frac{a_1}{d}\right) d \left (\frac{a_1}{d}+1 \right )d \left ( \frac{a_1}{d}+2 \right )\cdots d \left ( \frac{a_1}{d}+(n-1) \right ) \\[2pt]
&= d^n\prod_{k=0}^{n-1} \left(\frac{a_1}{d}+k\right)=d^n {\left(\frac{a_1}{d}\right)}^{\overline{n}}
\end{align}</math>
where <math>x^{\overline{n}}</math> denotes the [[Pochhammer symbol|rising factorial]].

By the recurrence formula <math>\Gamma(z+1)=z\Gamma(z)</math>, valid for a [[complex number]] <math>z>0</math>,

:<math>\Gamma(z+2)=(z+1)\Gamma(z+1)=(z+1)z\Gamma(z)</math>,

:<math>\Gamma(z+3)=(z+2)\Gamma(z+2)=(z+2)(z+1)z\Gamma(z)</math>,

so that

:<math> \frac{\Gamma(z+m)}{\Gamma(z)} = \prod_{k=0}^{m-1}(z+k)</math>

for <math>m</math> a positive integer and <math>z</math> a positive complex number.

Thus, if <math>a_1/d > 0 </math>,

:<math>\prod_{k=0}^{n-1} \left(\frac{a_1}{d}+k\right)= \frac{\Gamma{\left(\frac{a_1}{d} + n\right)}}{\Gamma{\left( \frac{a_1}{d} \right)}},</math>

and, finally,

:<math>a_1a_2a_3\cdots a_n = d^n\prod_{k=0}^{n-1} \left(\frac{a_1}{d}+k\right) = d^n \frac{\Gamma{\left(\frac{a_1}{d} + n\right)}}{\Gamma{\left( \frac{a_1}{d} \right)}} </math>

===Examples===

;Example 1
Taking the example <math> 3, 8, 13, 18, 23, 28, \ldots </math>, the product of the terms of the arithmetic progression given by <math>a_n = 3 + 5(n-1) </math> up to the 50th term is
:<math>P_{50} = 5^{50} \cdot \frac{\Gamma \left(3/5 + 50\right) }{\Gamma \left( 3 / 5 \right) } \approx 3.78438 \times 10^{98}. </math>

; Example 2
The product of the first 10 odd numbers <math>(1,3,5,7,9,11,13,15,17,19)</math> is given by
:<math> 1\cdot 3\cdot 5\cdots 19 =\prod_{k=0}^{9} (1+2k) = 2^{10} \cdot \frac{\Gamma \left(\frac{1}{2} + 10\right) }{\Gamma \left( \frac{1}{2} \right) } </math> = {{formatnum:654729075}}