Patterns

Arithmetic sequence

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2.

If the initial term of an arithmetic progression is $a_{1}$ and the common difference of successive members is $d$, then the $n$-th term of the sequence ($a_{n}$) is given by:

$\ a_{n}=a_{1}+(n-1)d$,

and in general

$\ a_{n}=a_{m}+(n-m)d$.

A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.

Sum

2	+	5	+	8	+	11	+	14	=	40
14	+	11	+	8	+	5	+	2	=	40
16	+	16	+	16	+	16	+	16	=	80

Computation of the sum 2 + 5 + 8 + 11 + 14. When the sequence is reversed and added to itself term by term, the resulting sequence has a single repeated value in it, equal to the sum of the first and last numbers (2 + 14 = 16). Thus 16 × 5 = 80 is twice the sum.

The sum of the members of a finite arithmetic progression is called an arithmetic series. For example, consider the sum:

$2+5+8+11+14$

This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2:

${\frac {n(a_{1}+a_{n})}{2}}$

In the case above, this gives the equation:

$2+5+8+11+14={\frac {5(2+14)}{2}}={\frac {5\times 16}{2}}=40.$

This formula works for any real numbers $a_{1}$ and $a_{n}$. For example:

$\left(-{\frac {3}{2}}\right)+\left(-{\frac {1}{2}}\right)+{\frac {1}{2}}={\frac {3\left(-{\frac {3}{2}}+{\frac {1}{2}}\right)}{2}}=-{\frac {3}{2}}.$

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:

$S_{n}=a_{1}+(a_{1}+d)+(a_{1}+2d)+\cdots +(a_{1}+(n-2)d)+(a_{1}+(n-1)d)$

$S_{n}=(a_{n}-(n-1)d)+(a_{n}-(n-2)d)+\cdots +(a_{n}-2d)+(a_{n}-d)+a_{n}.$

Adding both sides of the two equations, all terms involving d cancel:

$\ 2S_{n}=n(a_{1}+a_{n}).$

Dividing both sides by 2 produces a common form of the equation:

$S_{n}={\frac {n}{2}}(a_{1}+a_{n}).$

An alternate form results from re-inserting the substitution: $a_{n}=a_{1}+(n-1)d$:

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

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

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

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

Product

The product of the members of a finite arithmetic progression with an initial element a₁, common differences d, and n elements in total is determined in a closed expression

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

where $\Gamma$ denotes the Gamma function. The formula is not valid when $a_{1}/d$ is negative or zero.

This is a generalization from the fact that the product of the progression $1\times 2\times \cdots \times n$ is given by the factorial $n!$ and that the product

$m\times (m+1)\times (m+2)\times \cdots \times (n-2)\times (n-1)\times n$

for positive integers $m$ and $n$ is given by

${\frac {n!}{(m-1)!}}.$

Derivation

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

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

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

$\Gamma (z+2)=(z+1)\Gamma (z+1)=(z+1)z\Gamma (z)$,

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

so that

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

for $m$ a positive integer and $z$ a positive complex number.

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

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

and, finally,

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

Examples

Example 1

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

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

Example 2

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

$1.3.5\cdots 19=\prod _{k=0}^{9}(1+2k)=2^{10}\cdot {\frac {\Gamma \left({\frac {1}{2}}+10\right)}{\Gamma \left({\frac {1}{2}}\right)}}$ = 654,729,075

Standard deviation

The standard deviation of any arithmetic progression can be calculated as

$\sigma =|d|{\sqrt {\frac {(n-1)(n+1)}{12}}}$

where $n$ is the number of terms in the progression and $d$ is the common difference between terms. The formula is very similar to the standard deviation of a discrete uniform distribution.

Intersections

The intersection of any two doubly infinite arithmetic progressions is either empty or another arithmetic progression, which can be found using the Chinese remainder theorem. If each pair of progressions in a family of doubly infinite arithmetic progressions have a non-empty intersection, then there exists a number common to all of them; that is, infinite arithmetic progressions form a Helly family. However, the intersection of infinitely many infinite arithmetic progressions might be a single number rather than itself being an infinite progression.

Geometric sequence

Diagram illustrating three basic geometric sequences of the pattern 1(rⁿ⁻¹) up to 6 iterations deep. The first block is a unit block and the dashed line represents the infinite sum of the sequence, a number that it will forever approach but never touch: 2, 3/2, and 4/3 respectively.

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2.

Examples of a geometric sequence are powers r^k of a fixed non-zero number r, such as 2^k and 3^k. The general form of a geometric sequence is

$a,\ ar,\ ar^{2},\ ar^{3},\ ar^{4},\ \ldots$

where r ≠ 0 is the common ratio and a ≠ 0 is a scale factor, equal to the sequence's start value.

The distinction between a progression and a series is that a progression is a sequence, whereas a series is a sum.

Elementary properties

The n-th term of a geometric sequence with initial value a = a₁ and common ratio r is given by

$a_{n}=a\,r^{n-1}.$

Such a geometric sequence also follows the recursive relation

$a_{n}=r\,a_{n-1}$ for every integer $n\geq 2.$

Generally, to check whether a given sequence is geometric, one simply checks whether successive entries in the sequence all have the same ratio.

The common ratio of a geometric sequence may be negative, resulting in an alternating sequence, with numbers alternating between positive and negative. For instance

1, −3, 9, −27, 81, −243, ...

is a geometric sequence with common ratio −3.

The behaviour of a geometric sequence depends on the value of the common ratio.
If the common ratio is:

• positive, the terms will all be the same sign as the initial term.
• negative, the terms will alternate between positive and negative.
• greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term).
• 1, the progression is a constant sequence.
• between −1 and 1 but not zero, there will be exponential decay towards zero (→ 0).
• −1, the absolute value of each term in the sequence is constant and terms alternate in sign.
• less than −1, for the absolute values there is exponential growth towards (unsigned) infinity, due to the alternating sign.

Geometric sequences (with common ratio not equal to −1, 1 or 0) show exponential growth or exponential decay, as opposed to the linear growth (or decline) of an arithmetic progression such as 4, 15, 26, 37, 48, … (with common difference 11). This result was taken by T.R. Malthus as the mathematical foundation of his Principle of Population. Note that the two kinds of progression are related: exponentiating each term of an arithmetic progression yields a geometric progression, while taking the logarithm of each term in a geometric progression with a positive common ratio yields an arithmetic progression.

An interesting result of the definition of the geometric progression is that any three consecutive terms a, b and c will satisfy the following equation:

$b^{2}=ac$

where b is considered to be the geometric mean between a and c.

Geometric series

	2	+	10	+	50	+	250			=	312
− (			10	+	50	+	250	+	1250	=	5 × 312 )
	2							−	1250	=	(1 − 5) × 312

Computation of the sum 2 + 10 + 50 + 250. The sequence is multiplied term by term by 5, and then subtracted from the original sequence. Two terms remain: the first term, a, and the term one beyond the last, or ar^m. The desired result, 312, is found by subtracting these two terms and dividing by 1 − 5.

A geometric series is the sum of the numbers in a geometric progression. For example:

$2+10+50+250=2+2\times 5+2\times 5^{2}+2\times 5^{3}.$

Letting a be the first term (here 2), n be the number of terms (here 4), and r be the constant that each term is multiplied by to get the next term (here 5), the sum is given by:

${\frac {a(1-r^{n})}{1-r}}$

In the example above, this gives:

$2+10+50+250={\frac {2(1-5^{4})}{1-5}}={\frac {-1248}{-4}}=312.$

The formula works for any real numbers a and r (except r = 1, which results in a division by zero). For example:

$-2\pi +4\pi ^{2}-8\pi ^{3}=-2\pi +(-2\pi )^{2}+(-2\pi )^{3}={\frac {-2\pi (1-(-2\pi )^{3})}{1-(-2\pi )}}={\frac {-2\pi (1+2\pi ^{3})}{1+2\pi }}\approx -54.360768.$

Since the derivation (below) does not depend on a and r being real, it holds for complex numbers as well.

Derivation

To derive this formula, first write a general geometric series as:

$\sum _{k=1}^{n}ar^{k-1}=ar^{0}+ar^{1}+ar^{2}+ar^{3}+\cdots +ar^{n-1}.$

We can find a simpler formula for this sum by multiplying both sides of the above equation by 1 − r, and we'll see that

${\begin{aligned}(1-r)\sum _{k=1}^{n}ar^{k-1}&=(1-r)(ar^{0}+ar^{1}+ar^{2}+ar^{3}+\cdots +ar^{n-1})\\&=ar^{0}+ar^{1}+ar^{2}+ar^{3}+\cdots +ar^{n-1}-ar^{1}-ar^{2}-ar^{3}-\cdots -ar^{n-1}-ar^{n}\\&=a-ar^{n}\end{aligned}}$

since all the other terms cancel. If r ≠ 1, we can rearrange the above to get the convenient formula for a geometric series that computes the sum of n terms:

$\sum _{k=1}^{n}ar^{k-1}={\frac {a(1-r^{n})}{1-r}}.$

Related formulas

If one were to begin the sum not from k=1, but from a different value, say m, then

$\sum _{k=m}^{n}ar^{k}={\frac {a(r^{m}-r^{n+1})}{1-r}},$

provided $r\neq 1$. If $r=1$ then the sum is of just the constant $a$ and so equals $a(n-m+1)$.

Differentiating this formula with respect to r allows us to arrive at formulae for sums of the form

$G_{s}(n,r):=\sum _{k=0}^{n}k^{s}r^{k}.$

For example:

${\frac {d}{dr}}\sum _{k=0}^{n}r^{k}=\sum _{k=1}^{n}kr^{k-1}={\frac {1-r^{n+1}}{(1-r)^{2}}}-{\frac {(n+1)r^{n}}{1-r}}.$

For a geometric series containing only even powers of r multiply by  1 − r²  :

$(1-r^{2})\sum _{k=0}^{n}ar^{2k}=a-ar^{2n+2}.$

Then

$\sum _{k=0}^{n}ar^{2k}={\frac {a(1-r^{2n+2})}{1-r^{2}}}.$

Equivalently, take  r²  as the common ratio and use the standard formulation.

For a series with only odd powers of r

$(1-r^{2})\sum _{k=0}^{n}ar^{2k+1}=ar-ar^{2n+3}$

and

$\sum _{k=0}^{n}ar^{2k+1}={\frac {ar(1-r^{2n+2})}{1-r^{2}}}.$

An exact formula for the generalized sum $G_{s}(n,r)$ when $s\in \mathbb {N}$ is expanded by the Stirling numbers of the second kind as

$G_{s}(n,r)=\sum _{j=0}^{s}\left\lbrace {s \atop j}\right\rbrace x^{j}{\frac {d^{j}}{dx^{j}}}\left[{\frac {1-x^{n+1}}{1-x}}\right].$

Infinite geometric series

An infinite geometric series is an infinite series whose successive terms have a common ratio. Such a series converges if and only if the absolute value of the common ratio is less than one (|r| < 1). Its value can then be computed from the finite sum formula

$\sum _{k=0}^{\infty }ar^{k}=\lim _{n\to \infty }{\sum _{k=0}^{n}ar^{k}}=\lim _{n\to \infty }{\frac {a(1-r^{n+1})}{1-r}}={\frac {a}{1-r}}-\lim _{n\to \infty }{\frac {ar^{n+1}}{1-r}}$

Animation, showing convergence of partial sums of geometric progression $\sum \limits _{k=0}^{n}q^{k}$ (red line) to its sum ${1 \over 1-q}$ (blue line) for $|q|<1$.

Diagram showing the geometric series 1 + 1/2 + 1/4 + 1/8 + ⋯ which converges to 2.

Since:

$r^{n+1}\to 0{\mbox{ as }}n\to \infty {\mbox{ when }}|r|<1.$

Then:

$\sum _{k=0}^{\infty }ar^{k}={\frac {a}{1-r}}-0={\frac {a}{1-r}}$

For a series containing only even powers of $r$,

$\sum _{k=0}^{\infty }ar^{2k}={\frac {a}{1-r^{2}}}$

and for odd powers only,

$\sum _{k=0}^{\infty }ar^{2k+1}={\frac {ar}{1-r^{2}}}$

In cases where the sum does not start at k = 0,

$\sum _{k=m}^{\infty }ar^{k}={\frac {ar^{m}}{1-r}}$

The formulae given above are valid only for |r| < 1. The latter formula is valid in every Banach algebra, as long as the norm of r is less than one, and also in the field of p-adic numbers if |r|_p < 1. As in the case for a finite sum, we can differentiate to calculate formulae for related sums. For example,

${\frac {d}{dr}}\sum _{k=0}^{\infty }r^{k}=\sum _{k=1}^{\infty }kr^{k-1}={\frac {1}{(1-r)^{2}}}$

This formula only works for |r| < 1 as well. From this, it follows that, for |r| < 1,

$\sum _{k=0}^{\infty }kr^{k}={\frac {r}{\left(1-r\right)^{2}}}\,;\,\sum _{k=0}^{\infty }k^{2}r^{k}={\frac {r\left(1+r\right)}{\left(1-r\right)^{3}}}\,;\,\sum _{k=0}^{\infty }k^{3}r^{k}={\frac {r\left(1+4r+r^{2}\right)}{\left(1-r\right)^{4}}}$

Also, the infinite series 1/2 + 1/4 + 1/8 + 1/16 + ⋯ is an elementary example of a series that converges absolutely.

It is a geometric series whose first term is 1/2 and whose common ratio is 1/2, so its sum is

${\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots ={\frac {1/2}{1-(+1/2)}}=1.$

The inverse of the above series is 1/2 − 1/4 + 1/8 − 1/16 + ⋯ is a simple example of an alternating series that converges absolutely.

It is a geometric series whose first term is 1/2 and whose common ratio is −1/2, so its sum is

${\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{8}}-{\frac {1}{16}}+\cdots ={\frac {1/2}{1-(-1/2)}}={\frac {1}{3}}.$

Complex numbers

The summation formula for geometric series remains valid even when the common ratio is a complex number. In this case the condition that the absolute value of r be less than 1 becomes that the modulus of r be less than 1. It is possible to calculate the sums of some non-obvious geometric series. For example, consider the proposition

$\sum _{k=0}^{\infty }{\frac {\sin(kx)}{r^{k}}}={\frac {r\sin(x)}{1+r^{2}-2r\cos(x)}}$

The proof of this comes from the fact that

$\sin(kx)={\frac {e^{ikx}-e^{-ikx}}{2i}},$

which is a consequence of Euler's formula. Substituting this into the original series gives

$\sum _{k=0}^{\infty }{\frac {\sin(kx)}{r^{k}}}={\frac {1}{2i}}\left[\sum _{k=0}^{\infty }\left({\frac {e^{ix}}{r}}\right)^{k}-\sum _{k=0}^{\infty }\left({\frac {e^{-ix}}{r}}\right)^{k}\right]$.

This is the difference of two geometric series, and so it is a straightforward application of the formula for infinite geometric series that completes the proof.

Product

The product of a geometric progression is the product of all terms. It can be quickly computed by taking the geometric mean of the progression's first and last individual terms, and raising that mean to the power given by the number of terms. (This is very similar to the formula for the sum of terms of an arithmetic sequence: take the arithmetic mean of the first and last individual terms, and multiply by the number of terms.)

As the geometric mean of two numbers equals the square root of their product, the product of a geometric progression is:

$\prod _{i=0}^{n}ar^{i}=({\sqrt {a\cdot ar^{n}}})^{n+1}=({\sqrt {a^{2}r^{n}}})^{n+1}$.

(An interesting aspect of this formula is that, even though it involves taking the square root of a potentially-odd power of a potentially-negative r, it cannot produce a complex result if neither a nor r has an imaginary part. It is possible, should r be negative and n be odd, for the square root to be taken of a negative intermediate result, causing a subsequent intermediate result to be an imaginary number. However, an imaginary intermediate formed in that way will soon afterwards be raised to the power of $\textstyle n+1$, which must be an even number because n by itself was odd; thus, the final result of the calculation may plausibly be an odd number, but it could never be an imaginary one.)

Proof

Let P represent the product. By definition, one calculates it by explicitly multiplying each individual term together. Written out in full,

$P=a\cdot ar\cdot ar^{2}\cdots ar^{n-1}\cdot ar^{n}$.

Carrying out the multiplications and gathering like terms,

$P=a^{n+1}r^{1+2+3+\cdots +(n-1)+n}$.

The exponent of r is the sum of an arithmetic sequence. Substituting the formula for that calculation,

$P=a^{n+1}r^{\frac {n(n+1)}{2}}$,

which enables simplifying the expression to

$P=(ar^{\frac {n}{2}})^{n+1}=(a{\sqrt {r^{n}}})^{n+1}$.

Rewriting a as $\textstyle {\sqrt {a^{2}}}$,

$P=({\sqrt {a^{2}r^{n}}})^{n+1}$,

which concludes the proof.

Licensing

Content obtained and/or adapted from:

• Arithmetic progression, Wikipedia under a CC BY-SA license
• Geometric progression, Wikipedia under a CC BY-SA license