Derivation

Animated proof for the formula giving the sum of the first integers 1+2+...+n.

To derive the above formula, begin by expressing the arithmetic series in two different ways:

${\displaystyle S_{n}=a+a_{2}+a_{3}+\dots +a_{(n-1)}+a_{n}}$

${\displaystyle S_{n}=a+(a+d)+(a+2d)+\dots +(a+(n-2)d)+(a+(n-1)d).}$

Rewriting the terms in reverse order:

${\displaystyle S_{n}=(a+(n-1)d)+(a+(n-2)d)+\dots +(a+2d)+(a+d)+a.}$

Adding the corresponding terms of both sides of the two equations and halving both sides:

${\displaystyle S_{n}={\frac {n}{2}}[2a+(n-1)d].}$

This formula can be simplified as:

${\displaystyle {\begin{aligned}S_{n}&={\frac {n}{2}}[a+a+(n-1)d].\\&={\frac {n}{2}}(a+a_{n}).\\&={\frac {n}{2}}({\text{initial term}}+{\text{last term}}).\end{aligned}}}$

Furthermore, the mean value of the series can be calculated via: ${\displaystyle S_{n}/n}$:

${\displaystyle {\overline {a}}={\frac {a_{1}+a_{n}}{2}}.}$

The formula is very similar to the mean of a discrete uniform distribution.

Derivation

${\displaystyle {\begin{aligned}a_{1}a_{2}a_{3}\cdots a_{n}&=\prod _{k=0}^{n-1}(a_{1}+kd)\\&=\prod _{k=0}^{n-1}d\left({\frac {a_{1}}{d}}+k\right)=d\left({\frac {a_{1}}{d}}\right)d\left({\frac {a_{1}}{d}}+1\right)d\left({\frac {a_{1}}{d}}+2\right)\cdots d\left({\frac {a_{1}}{d}}+(n-1)\right)\\&=d^{n}\prod _{k=0}^{n-1}\left({\frac {a_{1}}{d}}+k\right)=d^{n}{\left({\frac {a_{1}}{d}}\right)}^{\overline {n}}\end{aligned}}}$

where ${\displaystyle x^{\overline {n}}}$ denotes the rising factorial.

By the recurrence formula ${\displaystyle \Gamma (z+1)=z\Gamma (z)}$, valid for a complex number ${\displaystyle z>0}$,

${\displaystyle \Gamma (z+2)=(z+1)\Gamma (z+1)=(z+1)z\Gamma (z)}$,

${\displaystyle \Gamma (z+3)=(z+2)\Gamma (z+2)=(z+2)(z+1)z\Gamma (z)}$,

so that

${\displaystyle {\frac {\Gamma (z+m)}{\Gamma (z)}}=\prod _{k=0}^{m-1}(z+k)}$

for ${\displaystyle m}$ a positive integer and ${\displaystyle z}$ a positive complex number.

Thus, if ${\displaystyle a_{1}/d>0}$,

${\displaystyle \prod _{k=0}^{n-1}\left({\frac {a_{1}}{d}}+k\right)={\frac {\Gamma \left({\frac {a_{1}}{d}}+n\right)}{\Gamma \left({\frac {a_{1}}{d}}\right)}}}$,

and, finally,

${\displaystyle a_{1}a_{2}a_{3}\cdots a_{n}=d^{n}\prod _{k=0}^{n-1}\left({\frac {a_{1}}{d}}+k\right)=d^{n}{\frac {\Gamma \left({\frac {a_{1}}{d}}+n\right)}{\Gamma \left({\frac {a_{1}}{d}}\right)}}}$

Examples

Example 1

Taking the example ${\displaystyle 3,8,13,18,23,28,\ldots }$, the product of the terms of the arithmetic progression given by ${\displaystyle a_{n}=3+5(n-1)}$ up to the 50th term is

${\displaystyle P_{50}=5^{50}\cdot {\frac {\Gamma \left(3/5+50\right)}{\Gamma \left(3/5\right)}}\approx 3.78438\times 10^{98}.}$

Example 2

The product of the first 10 odd numbers ${\displaystyle (1,3,5,7,9,11,13,15,17,19)}$ is given by

${\displaystyle 1\cdot 3\cdot 5\cdots 19=\prod _{k=0}^{9}(1+2k)=2^{10}\cdot {\frac {\Gamma \left({\frac {1}{2}}+10\right)}{\Gamma \left({\frac {1}{2}}\right)}}}$ = 654,729,075