Quadratic Sequence

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In mathematics, a sequence is an ordered list of numbers or objects in which each element is uniquely identified by its position or index. Similarly, the term quadratic describes something that pertains to squares, to the operation of squaring, to terms of the second degree, or equations or formulas that involve such terms.

To sum up, a quadratic sequence is a sequence of numbers in which the difference between consecutive terms is not constant, rather difference of the second differences are equal and follows a quadratic pattern. The highest degree of the general term of a quadratic sequence is 2. In a quadratic sequence, the terms are generated by a quadratic equation, which is a polynomial equation of degree 2. The general formula of a quadratic sequence can be expressed as:

T_n = $ax^{2}+bx+c$

Where T_n represents the n^th term of the sequence, and a, b, and c are constants that determine the specific values of the sequence.

Example of a quadratic sequence:

The general term of quadratic sequence can be expressed in different ways. In this article, we will be looking into two of those.

Quadratic sequence

General term of quadratic sequence

First Derivation

$1$$st$ $term(a)=a$
$2$$nd$ $term$ ($a$$1$) $=a+b$
$3$$rd$ $term$ ($a$$2$) $=a$$1$$+b$$1$ = $a$+$b$+$b$$1$
$4$$th$ $term$ ($a$$3$) $=a$$2$$+b$$2$ $=a$+$b$+$b$$1$$+b$$2$
$n$$th$ $term=$ $a$$n-2$$+b$$n-2$ $=a$+$b$+$b$$1$$+b$$2$$+...+b$$n-1$

$Since$ [ $b$+$b$_$1$$+b$_$2$$+...+b$_$n-1$ ] is an arithmetic sequence with a common difference (c), using the formula of sum of terms of arithmetic sequence:

$Sn={\frac {n}{2}}[2a+(n-1)d]$, where:

$number$ $of$ $terms$ $(n)$ $=n-1,$

$first$ $term$ $(a)=b,$ $difference$ $(d)=c$

$b$+$b$$1$$+b$$2$$+...+b$$n-1$$=$  ${\frac {n-1}{2}}$ $[2b+(n-2)c]$ 

$n$^$th$ $term=$ $a+$${\frac {1}{2}}$ $(n-1)$$(nc-2c+2b)$

Hence, general term of the quadratic sequence is:

(t_n) = ${\frac {1}{2}}$$(n-1)$$(nc-2c+2b)$$+a$

where a is the first term of the first sequence, b is the first term of the sequence formed by the differences of the first sequence, and c is the difference of the terms of the second sequence formed by the differences of the first sequence.

Examples

Example 1

Taking an example of a quadratic sequence: $8,14,22,32,44,58,...$

First term of the first sequence is 8, which is: a=8

Sequence formed by the differences of the consecutive terms of the first sequence is: 6, 8, 10, 12, 14, ... So, first term of the second sequence is 6, which is: b=6 and constant difference of the second sequence is 2, which is: c=2

Putting a=8, b=6, and c=2 in $tn=$ ${\frac {1}{2}}$ $(n-1)$$(nc-2c+2b)$$+a$:

$tn=$ ${\frac {1}{2}}*(n-1)(2n-2*2+2*6)+8$

$tn=$=${\frac {1}{2}}*(n-1)(2n+8)+8$

Therefore, $tn=$ =$n^{2}+3n+4$

Example 2

Taking another example of a quadratic sequence: $82,87,101,124,156,197,...$

a=82,

Sequence formed by the differences of the consecutive terms of the first sequence is: 5, 14, 23, 32, 41, ... So,

b=5, c= 9

putting the values of a, b, c in $tn=$ ${\frac {1}{2}}$ $(n-1)$$(nc-2c+2b)$$+a$ :

$tn=$ ${\frac {1}{2}}*(n-1)(9n-2*9+2*5)+82$

$tn=$ =${\frac {1}{2}}*(n-1)(9n-8)+8$2

Therefore ,$tn=$ ${\frac {1}{2}}(9n^{2}-17n+43)$

Second Derivation

t0 = $a$
t1 = t0 + b = $a+b$
t2 = t1 + b1 = $a+b+b+c=a+2b+c$
t3 = t2 + b2 = $a+2b+c+b+c+c=a+3b+3c$
t4 = t3 + b3 = $a+3b+3c+b+c+c+c=a+4b+6c$
t5 = t4 + b4 $=a+4b+6c+b+c+c+c+c=a+5b+10c$

By analyzing the coefficients, the following things can be concluded:

• Each and every term has only one a (first term).
• The Coefficients of b are in the following sequence: 0, 1, 2, 3, 4, 5, ...,n. Which can be obtained by the formula: $n$.
• The coefficients of c are in the following sequence: 0, 0, 1, 3, 6, 10....n. Which can be obtained by the formula: ${\frac {n(n-1)}{2}}$

Now, from the above three points, we conclude the following formula:

$tn=a+bn+c{\frac {n(n-1)}{2}}$

In this formula:

first term = t0
second term = t1
third term = t2
nth term = tn-1

In order to reduce the complexity of the formula, the formula is revised where $n=n-1$:

So, general term of the quadratic sequence is also:

$tn=a+b(n-1)+c{\frac {(n-1)(n-2)}{2}}$

where a is the first term of the first sequence, b is the first term of the sequence formed by the differences of the first sequence, and c is the difference of the terms of the second sequence formed by the differences of the first sequence.

Also in this formula:

first term = t1
second term = t2
third term = t3
nth term = tn

Examples

Example 1

Taking an example of a quadratic sequence: $8,14,22,32,44,58,...$

First term of the first sequence is 8, which is: a=8

Sequence formed by the differences of the consecutive terms of the first sequence is: 6, 8, 10, 12, 14, ... So, first term of the second sequence is 6, which is: b=6 and constant difference of the second sequence is 2, which is: c=2

Putting a=8, b=6, and c=2 in $tn=a+b(n-1)+c{\frac {(n-1)(n-2)}{2}}$:

$tn=8+6(n-1)+2{\frac {(n-1)(n-2)}{2}}$

$tn=8+6n-6+(n-1)(n-2)$

$tn=2+6n+n^{2}-3n+2$

Therefore, $tn=n^{2}+3n+4$

Example 2

Taking another example of a quadratic sequence: $82,87,101,124,156,197,...$

a=82,

Sequence formed by the differences of the consecutive terms of the first sequence is: 5, 14, 23, 32, 41, ... So,

b=5, c= 9

Putting the values of a, b, c in $tn=a+b(n-1)+c{\frac {(n-1)(n-2)}{2}}$

$tn=82+5(n-1)+9{\frac {(n-1)(n-2)}{2}}$

$tn=82+5n-5+9{\frac {(n^{2}-3n+2)}{2}}$

$tn=77+5n+{\frac {9n^{2}}{2}}-{\frac {27n}{2}}+{\frac {9*2}{2}}$

$tn={\frac {9n^{2}}{2}}-{\frac {17n}{2}}+86$

Therefore, $tn={\frac {1}{2}}(9n^{2}-17n+43)$

Sum of terms of quadratic sequence

General formula of the sum of terms of quadratic sequence

Using the general formula of the quadratic sequence found from Second Derivation:

$tn=a+b(n-1)+c{\frac {(n-1)(n-2)}{2}}$

Sum of terms:

Let the number of terms (n) be 'k', in order to avoid contradiction.

nΣk = 1t_k = nΣk = 1$a+b$nΣk = 1$(k-1)$ +${\frac {c}{2}}$nΣk = 1$(k-1)(k-2)$

= $an$+ $b$ ${\frac {n}{2}}(n-1)$ + ${\frac {c}{2}}$*${\frac {1}{3}}n(n-1)(n-2)$

= $an$ + ${\frac {n}{2}}(n-1)$$b+{\frac {c(n-2)(n-1)}{6}}$

= $an$ + ${\frac {n}{2}}(n-1)$$[b+{\frac {c(n-2)}{3}}]$

=$an$ + ${\frac {n}{2}}(n-1)$${\frac {(3b+cn-2c)}{3}}$

Hence, the general formula for the sum of the terms of quadratic sequence is:

$S$$n$ $=$ ${\frac {n}{6}}(n-1)$$(3b+cn-2c)$ $+an$

where a is the first term of the first sequence, b is the first term of the sequence formed by the differences of the first sequence, and c is the difference of the terms of the second sequence formed by the differences of the first sequence.

Examples

Example 1

Taking an example of a quadratic sequence: $1,2,4,7,11.....$

Let's find the sum of terms from 1 to 5.

a = 1, b=1, c=1

number of terms (n) = 5

Putting the values of a, b, c, n in $S$$n$ $=$ ${\frac {n}{6}}(n-1)$$(3b+cn-2c)$ $+an$

$S$₅ = ${\frac {5}{6}}(5-1)(3*1+1*5-2*1)+1*5$

=${\frac {5}{6}}*4*6+5$

= $25$

Hence, the sum of the terms of the quadratic sequence ($1,2,4,7,11.....$) from 1 to 5 is 25.

Example 2

Taking an example of a quadratic sequence: $6,10,17,27,40....$

Let's find the sum of terms from 1 to 50.

a = 6, b=4, c=3

number of terms (n) = 50

Putting the values of a, b, c, n in $S$$n$ $=$ ${\frac {n}{6}}(n-1)$$(3b+cn-2c)$ $+an$

$S$₅₀ = ${\frac {50}{6}}(50-1)(3*4+3*50-2*3)+6*50$

= ${\frac {50}{6}}*49*156+300$

= $64000$

Hence, the sum of the terms of the quadratic sequence ($6,10,17,27,40....$) from 1 to 50 is 64000.

Geometric Quadratic Sequence

A geometric quadratic sequence is a sequence of numbers in which the ratio between consecutive terms follows a quadratic pattern. In this sequence, the ratio of the ratios is equal. This is an example of a geometric quadratic sequence.

Example of geometric quadratic sequence.

General term of geometric quadratic sequence

t1 = $a$
t2 = t1 $*b$ =  $ab$
t3 = t2 $*b*c$  =  $ab*b*c$  = $ab^{2}c$
t4 = t3 $*b*c*c$  =  $ab^{2}c$ $*b*c*c$  = $ab^{3}c^{3}$
t5 = t4 $*b*c*c*c$  =  $ab^{3}c^{3}$ $*b*c*c*c$ = $ab^{4}c^{6}$
t6 = t5 $*b*c*c*c*c$ = $ab^{4}c^{6}$ $*b*c*c*c*c$ = $ab^{5}c$10

By analyzing the powers of a, b, c, the following things can be concluded:

• Each and every term has only one a (first term).
• Powers of b can be written as $(n-1)$.
• powers of c are in the quadratic sequence: 0,0,1,3,6,10. And general term of it is ${\frac {1}{2}}(n-1)(n-2)$.

Now, from the above three points:

Suvanjan's formula for the general term of geometric quadratic sequence is:

$tn=$ $a*$ $b$^$n-1$ $*c$ ^{$1/2*(n-1)(n-2)$}

where a is the first term of the first sequence, b is the first term of the sequence formed by the ratios of the first sequence, and c is the ratio of the terms of second sequence formed by the ratio of the first sequence.

Examples

Example 1

Taking an example of a geometric quadratic sequence: $6,18,108,1296,31104....$

a = 6, b=3, c=2

Putting the values of a, b, c in $tn=$ $a*$ $b$^$n-1$ $*c$ ^{$1/2*(n-1)(n-2)$}

$tn=$ $6*$ 3^$n-1$ $*2$ ^{$1/2*(n-1)(n-2)$}

$tn=$$3*2*$ 3^$n-1$ $*2$ ^{$1/2*(n-1)(n-2)$}

$tn=$ 3^$n-1+1$ $*2$ ^{$1/2*(n-1)(n-2)+1$}

Therefore, $tn=$3^$n$ $*2$ ^{$1/2*(n-1)(n-2)+1$}

Example 2

Taking an example of a geometric quadratic sequence: $2,14,294,18522,....$

a = 2, b=7, c=3

Putting the values of a, b, c in $tn=$ $a*$ $b$^$n-1$ $*c$ ^{$1/2*(n-1)(n-2)$}

Therefore, $tn=$ $2*$ 7^$n-1$ $*3$ ^{$1/2*(n-1)(n-2)$}

References

1. Sum of terms of arithmetic sequence: Arithmetic progression - Wikipedia
2. Introduction
3. Overview: quadratic sequence
4. Quadratic patterns
5. https://ebeducationservices.co.uk/wp-content/uploads/2019/08/EB-How-to-Work-with-Sequences-Part-2.pdf
6. Outcomes and Assessment Standards

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