Establishment of Kifilideen coefficient tables for positive and negative powers of \(n\) and \(-n\) of Kifilideen trinomial theorem and other development based on matrix and standardized methods

1. Introduction

Goss [1] and Aljohani [2] indicate that the Binomial theorem for negative and fraction powers of \(-n\) and \(\frac{a}{b}\) were developed by Sir Isaac Newton (I642 – 1727) in 1665. Bombelli (1572) gave the coefficients of the binomial expansion of \({\left(a+b\right)}^n\ \)for \(n\ =\ 1,\ 2,\ 3,\ 4,\dots ,\ 7\) and Oughtred (1631) provided them for \(n\ =\ 1\ ,\ 2,\ 3,\ 4,\ 5,\dots ,\ 10\) [3-5]. Blaise Pascal (1623 – 1662), a French mathematician, developed the Pascal triangle, also known as the Yanghui triangle, in 1664 to generate the coefficient of terms of the positive power of binomial theorem [6-10]. Although, from the triangle, there is no indication of how the coefficients are progressing from one term to the other. However, from the binomial theorem expansion, it could be deduced which term owns the coefficient in the Pascal triangle. No publication is available for coefficients of negative power of \(-n\) either in triangle form or any other form. This delays the process of generating the series of the binomial theorem involving the negative power of \(-n\).

A theorem of Kifilideen matrix transformation of the positive power of \(n\) of trinomial expression in which three variables \(x,y\), and \(z\) are found in parts of the trinomial expression was originated. The development would ease evaluating the trinomial expression’s positive power of \(n\).

Kifilideen (2020) developed the Kifilideen trinomial theorem of the positive power of \(n\) for the expansion of the form \({\left[x+y+z\right]}^n\) using the matrix approach, which helps in arranging the terms of the expansion in an orderly and standardized manner and ways [11]. It was observed that the process of generating the coefficient of each term of the expansion of positive and negative powers of \(n\) and \(-n\) of the trinomial theorem is stressful and time-consuming which the same problem is identified with coefficients of terms of a binomial theorem of powers of \(n\) and \(-n\) [12,13]. This slows the process of producing the series of any particular trinomial expansion.

The establishment of tables of coefficients of positive and negative powers of binomial and trinomial theorems would help and be handy in generating the coefficient of each term of a series and result in easy expansion. In the Tables, each term and its corresponding coefficient will be indicated and linked together. Also, it would show how the coefficients are progressing from one term to the other. This will remove the stress that is been encountered in the expansion process if the Tables are fully utilized. It can serve as a guide and would also help in easy visualization of patterns and analysis in which the coefficients are progressing for each power of \(n\) of the binomial and trinomial theorems. Kifilideen matrix approach had been used to evaluate and compute the power of base of eleven, other bi-digits, and tri-digit numbers [14-18]. This study established Kifilideen coefficient tables for positive and negative powers of \(n\) and \(-n\) for the Kifilideen trinomial theorem.

2. Materials and Methods

2.1. Pascal coefficient table for terms of series of expansion of positive power of \(n\) of Newton’s binomial theorem

Table 1 indicates Pascal coefficient table for terms of series of expansion of positive power of \(n\) of Newton’s binomial theorem. The series of the terms of Newton’s binomial theorem for a particular positive power of \(n\) is finite.

2.2. Kifilideen coefficient table for negative power of \(-n\) of Newton’s binomial theorem

Table 2 shows Kifilideen coefficient table for negative power of \(-n\) of Newton’s binomial theorem. The series of Newton’s expansion of negative power of \(-n\) of binomial theorem gives infinite series which can be shown in the Table below. The series of the terms of Newton’s binomial theorem for a particular negative power of \(n\) is infinite.

2.3. Kifilideen power combination table of the terms of the series of Kifilideen expansion of positive power of n of trinomial expression \((x+y+z)\) in a standardized order

Table 3 presents the Kifilideen power combination table of the terms of the series of Kifilideen expansion of positive power of \(n\) of trinomial expression \((x+y+z)\) in a standardized order. Each degree of the positive power of \(n\) of the Kifilideen expansion of the Kifilideen trinomial theorem has a finite power combination in the series.

2.4. Kifilideen coefficient table for positive power of n of Kifilideen trinomial theorem

Table 4 presents the Kifilideen coefficient table for positive power of \(n\) of Kifilideen trinomial theorem based on matrix approach. The series of Kifilideen expansion of positive power of \(n\) of trinomial theorem gives finite series which can be shown in the Table below. From the Table 4; the change in colour from blue to red then to blue and so on till the last term for any particular positive power of \(n\) indicates the migration from one group to another. Also, from the Table 4, the maximum number of terms generated by any positive power of\(\ n\) can be known. The formula to determine the maximum number of terms generated by any positive power of \(n\) of a trinomial theorem is presented in the Kifilideen (2020) on the publication of development of Kifilideen trinomial theorem using matrix approach [11].

2.5. Kifilideen coefficient table for negative power of -n of Kifilideen trinomial theorem based on matrix and standardized approach

Table 5 shows Kifilideen coefficient table for negative power of \(-n\) of Kifilideen trinomial theorem based on matrix and standardized approach. The series of Kifilideen expansion of negative power of \(-n\) of trinomial theorem gives infinite series which can be shown in the Table 5 below. From the Table 5; the change in colour from blue to red then to blue and so on to infinity for any particular negative power of \(-n\) indicates the migration from one group to another. A clear study of Table 5 indicates that the coefficients of the negative power of \(- 1\) of Kifilideen trinomial theorem in periodicity and orderly manner can be generated from coefficients of Pascal triangle of binomial theorem of positive power of\(\ n\). The coefficients of the negative power of \(-1\) of Kifilideen trinomial theorem in orderly manner are \(1\); \(-\)1, \(-\)1; \(1\), \(2\), \(1\); \(-\)1, \(-\)3, \(-\)3, \(-\)1; \(1\), \(4\), \(6\), \(4\), \(1\); \(-\)1, \(-\) 5, \(-\)10, \(-\ 10\), \(-\)5, \(-\ \)1; \(1\), \(6\), \(15\), \(20\), \(15\), \(6\), \(1\);….

Also, the study of the Table 5 reveals that the coefficient of the negative power of \(-2\) of Kifilideen trinomial theorem in periodicity and orderly manner can be generated from the coefficients of Pascal triangle of binomial theorem of positive power of\(\ n\). The coefficients of the negative power of \(-2\) of Kifilideen trinomial theorem in orderly manner are\(\ 1\) (\(\times 1\)); \(-\)1, \(-\)1(\(\times 2\)); 1, 2, 1 (\(\times 3\)); \(-\)1, \(-\)3, \(-\)3, \(-\)1 (\(\times 4\)); \(1\), \(4\), \(6\), \(4\),\(\ 1\) (\(\times 5\)); \(-\)1, \(-\) 5, \(-\)10, \(-\) 10, \(-\)5, \(-\ \)1 (\(\times 6\)); \(1\), \(6\), \(15\), \(20\), 15, \(6\), \(1\) (\(\times 7\));… . So, the coefficients of the negative power of \(-2\) of Kifilideen trinomial theorem in orderly manner are \(1\); \(-2\), \(-2\); \(3\), \(6\), \(3\); \(-\)4, \(-\)12, \(-\)12, \(-\)4; \(5\), 20, 30, 20, 5; \(-\)6, \(-\) \(30\), \(-60\), \(-\) 60, \(-\)30, \(-\ \)6; 7, \(42\), 105, \(140\), \(105\), 42, \(7\);… .

2.6. Kifilideen theorem of matrix transformation of positive power of \(n\) of trinomial expression

If three variables \(x, y\)and \(z\) are found in each part of trinomial expression of positive power of \(n\) such as \[\ {\ \left[ux^ay^bz^c+vx^dy^ez^g+wx^hy^mz^p\right]}^n \tag{1}\]

and the power combination of any term in the Kifilideen expansion of that kind of positive power of the trinomial expression is set as \(kif\) while the value of this term is designated as \(qx^ry^sz^t\).

Then, the kifilideen matrix transformation of such positive power of \(n\) of the trinomial expression is of the form;

\[ \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]: \left[ \begin{array}{ccc} a & d & h \\ b & e & m \\ c & g & p \end{array} \right] \left[ \begin{array}{c} k \\ i \\ f \end{array} \right]=\left[ \begin{array}{c} r \\ s \\ t \end{array} \right], \tag{2}\]

Thus \(k+i+f=n\) and where \(u,v,\) \(w\ \)and \(q\ \)are constants, More so, \[\ {}^{-n}_{kif}{Cu^k}v^iw^f=q, \tag{3}\]

3. Results

3.1 Utilization of Kifilideen coefficient table for Kifilideen trinomial theorem based standardized and matrix approach

[i] Expand the following using Kifilideen coefficient table

[a] \({\left[x+y+z\right]}^7\) [b] \({\mathrm{\ }\left(\frac{x^2}{y}-yz^3+xz\right)}^4\) [c] \({\mathrm{\ }\left(x+y+z\right)}^{-\ 5}\) [d] \({\mathrm{\ }\left(1+x+y\right)}^{-\ 3}\)

Solution

[a] From Kifilideen coefficient Table 3 for positive power of \(7\), the coefficienst in ascending order are \(1\),\(7\),\(21\),\(35\), \(35,\ 21\),\(7\),\(1\),7,42,105,140,105,42,7,21,105,210,210,105,21,35,140,210,105,21,35,140,210,140,35,105,105,35,21,42,\(21\), \(7\),\(7\),\(1\). So the expansion of \({\left[x+y+z\right]}^7\ \)using Kifilideen trinomial theorem based on standardized and matrix approach, we have:

\[{\left[x+y+z\right]}^7 = 1 \times x^7\times y^0\times z^0+\mathrm{7\ }\times x^6\times y^1\times z^0 +\mathrm{21}\times x^5\times y^2\times z^0+\mathrm{35\ }\times x^4\times y^3\times z^0+\mathrm{35\ }\times x^3\times y^4\times z^0\]

\[+\mathrm{21\ }\times x^2\times y^5\times z^0+\mathrm{7\ }\times x^1\times y^6\times z^0+\mathrm{1\ }\times x^0\times y^7\times z^0 +\mathrm{7\ }\times x^6\times y^0\times z^1+\mathrm{42\ }\times x^5\times y^1\times z^1+\mathrm{105\ }\]

\[\times x^4\times y^2\times z^1+\mathrm{140\ }\times x^3\times y^3\times z^1+\mathrm{105\ }\times x^2\times y^4\times z^1\mathrm{+42\ }\times x^1\times y^5\times z^1+\mathrm{7\ }\times x^0\times y^6\times z^1\]

\[+21\times x^5{\times y}^0{\times z}^2+105\mathrm{\ }x^4{\times y}^1{\times z}^2+\mathrm{210\ }x^3{\times y}^2{\times z}^2+210\mathrm{\ }x^2{\times y}^3{\times z}^2+\mathrm{105\ }x^1{\times y}^4{\times z}^2+2\mathrm{1\ }x^0{\times y}^5{\times z}^2+\mathrm{35\ }\]

\[\times x^4{\times y}^0{\times z}^3+140\mathrm{\ }\times x^3{\times y}^1{\times z}^3+210\mathrm{\ }\times x^2{\times y}^2{\times z}^3+\mathrm{140\ }\times x^1{\times y}^3{\times z}^3\]

\[+\times x^0{\times y}^4{\times z}^3 +\mathrm{35\ }\times x^3{\times y}^0{\times z}^4+\mathrm{105\ }\times x^2{\times y}^1{\times z}^4 +\mathrm{105\ }\times x^1{\times y}^2{\times z}^4 +\mathrm{35\ }\times x^0{\times y}^3{\times z}^4+2\mathrm{1\ }\times x^2{\times y}^0{\times z}^5\]

\[\label{GrindEQ__4_}+\mathrm{42\ }\times x^1{\times y}^1{\times z}^5+2\mathrm{1\ }\times x^0{\times y}^2{\times z}^5+\mathrm{7\ }\times x^1{\times y}^0{\times z}^6+\mathrm{7\ }\times x^0{\times y}^1{\times z}^6+\mathrm{1\ }\times x^0{\times y}^0{\times z}^7, \tag{4}\]

\[{\left[x+y+z\right]}^7 = x^7+\mathrm{7\ }x^6y +2\mathrm{1\ }x^5y^2+\mathrm{35\ }x^4y^3+\mathrm{35\ }x^3y^4+\mathrm{21\ }x^2y^5+\mathrm{7\ }xy^6+y^7 +\mathrm{7\ }\times x^6z+\mathrm{42\ }x^5yz+\mathrm{105\ }x^4y^2z\]

\[+\mathrm{140\ }x^3y^3z+\mathrm{105\ }x^2y^4z\mathrm{+42\ }xy^5z+\mathrm{7\ }y^6z+21x^5z^2+105\mathrm{\ }x^4yz^2+\mathrm{210\ }x^3y^2z^2+210\mathrm{\ }x^2y^3z^2+\mathrm{105\ }xy^4z^2+2\mathrm{1\ }y^5z^2\] \[+\mathrm{35\ }x^4z^3+140\mathrm{\ }x^3yz^3+210\mathrm{\ }x^2y^2z^3+\mathrm{140\ }xy^3z^3+\mathrm{35\ }y^4z^3+\mathrm{35\ }x^3z^4+\mathrm{105\ }x^2y^1z^4 +\mathrm{105\ }xy^2z^4 +\mathrm{35\ }y^3z^4+2\mathrm{1\ }x^2z^5\] \[\ +\mathrm{42\ }xyz^5+2\mathrm{1}y^2z^5+\mathrm{7}xz^6+\mathrm{7\ }yz^6+z^7, \tag{5}\]

[b] From Kifilideen coefficient Table 3 for positive power of 4, the coefficients in ascending order are 1,4,6,4,1,4,12,12,4,6,12,6,4,4,1. So the expansion of \({\mathrm{\ }\left(\frac{x^2}{y}-yz^3+xz\right)}^4\ \)using Kifilideen trinomial theorem of positive power of \(n\) based on standardized and matrix approach, we have:

\[{\left(x^2y^{-1}-yz^3+xz\right)}^4=1{\times \left[x^2y^{-1}\right]}^4{\left[-yz^3\right]}^0{\left[xz\right]}^0{+4\left[x^2y^{-1}\right]}^3{\left[-yz^3\right]}^1{\left[xz\right]}^0{+6\left[x^2y^{-1}\right]}^2{\left[-yz^3\right]}^2{\left[xz\right]}^0+\] \[{+4\left[x^2y^{-1}\right]}^1{\left[-yz^3\right]}^3{\left[xz\right]}^0+{\left[x^2y^{-1}\right]}^0{\left[-yz^3\right]}^4{\left[xz\right]}^0+4{\left[x^2y^{-1}\right]}^3{\left[-yz^3\right]}^0{\left[xz\right]}^1{+12\left[x^2y^{-1}\right]}^2{\left[-yz^3\right]}^1{\left[xz\right]}^1\] \[\ {+12\left[x^2y^{-1}\right]}^1{\left[-yz^3\right]}^2{\left[xz\right]}^1+{4\left[x^2y^{-1}\right]}^0{\left[-yz^3\right]}^3{\left[xz\right]}^1{+6\left[x^2y^{-1}\right]}^3{\left[-yz^3\right]}^0{\left[xz\right]}^2{+12\left[x^2y^{-1}\right]}^2{\left[-yz^3\right]}^1{\left[xz\right]}^2\] \[\ {+6\left[x^2y^{-1}\right]}^1{\left[-yz^3\right]}^2{\left[xz\right]}^2{+4\left[x^2y^{-1}\right]}^1{\left[-yz^3\right]}^0{\left[xz\right]}^3{+4\left[x^2y^{-1}\right]}^0{\left[-yz^3\right]}^1{\left[xz\right]}^3+{\left[x^2y^{-1}\right]}^0{\left[-yz^3\right]}^0{\left[xz\right]}^4, \tag{6}\]

\[{\mathrm{\ }\left(x^2y^{-1}-yz^3+xz\right)}^4=x^8y^{-4}-4x^6y^{-2}z^3+{6x}^4z^6-{4x}^2y^2z^9+y^4z^{12}{+4x}^7y^{-3}z-12{x^5y}^{-1}z^4+{12x}^3yz^7\] \[\ -4xy^3z^{10}+{6x}^8y^{-3}z^2-{12x}^4y^{-1}z^5+{6x}^4yz^8+{4x}^5y^{-1}z^3-{4x}^3yz^6+x^4z^4, \tag{7}\] [c] From Kifilideen coefficient Table 4 for negative power of \(-5\), the coefficients in ascending order are \(1, -5, -5, 15, 30, 15, -35, -105, -105,-35, 70, 280, 420, 280, 70, -126, -630, -1260, -1260, -630, -126,\)
\(210, 1260, 3150, 4200,3150, 1260, 210, —\) . The expansion gives infinite series. So the expansion of\({\mathrm{\ }\left(x+y+z\right)}^{-\ 5}\) using Kifilideen trinomial theorem of negative power of – n based on standardized and matrix approach, we have:

\[{\left(x+y+z\right)}^{-\ 5}= x^{-5}y^0z^0 — 5x^{-6}y^1z^0-\mathrm{5}x^{-6}y^0z^1+15x^{-7}y^2z^0 +30x^{-7}y^1z^1+15x^{-7}y^0z^1-35x^{-8}y^3z^0-\] \[105x^{-8}y^2z^1-105x^{-8}y^1z^2-{35x}^{-8}y^0z^3+{70x}^{-9}y^4z^0+{280x}^{-9}y^3z^1+{420x}^{-9}y^2z^2{+280x}^{-9}y^1z^3\] \[+{70x}^{-9}y^0z^4{-126x}^{-10}y^5z^0{-630x}^{-10}y^4z^1{-1260x}^{-10}y^3z^2{-1260x}^{-10}y^2z^3{-630x}^{-10}y^1z^4{-126x}^{-10}y^0z^5\] \[{+210x}^{-11}y^6z^0{+1260x}^{-10}y^5z^1{+3150x}^{-10}y^4z^2{+4200x}^{-10}y^3z^3{+3150x}^{-10}y^2z^4{+1260x}^{-10}y^1z^5{+210x}^{-11}y^0z^6\]

[d] From Kifilideen coefficient table 4 for negative power of -3, the coefficients in ascending order are \(1, – 3, -3, 6, 12, 6, -10, -30, -30, -10, 15, 60, 90, 60, 15, -21, -105, -210, -210, -105, -21, 28, 168, 420, 560, 420,\)
\(168, 28, —-\). The expansion gives infinite series. So the expansion of\({\mathrm{\ }\left(1+y+z\right)}^{-\ 3}\) using Kifilideen trinomial theorem of negative power of \(\ -\ n\ \) based on standardized and matrix approach, we have:

\[{\mathrm{\ }\left(1+x+y\right)}^{-\ 3}=1-3x^1y^0-3x^0y^1+6x^2y^0+12x^1y^1+6x^0y^2-10x^3y^0-30x^2y^1-30x^1y^2-10x^0y^3\] \[+15x^4y^0+60x^3y^1+90x^2y^2+6{0x}^1y^3+15x^0y^4{-21y}^5z^0{-105y}^4z^1{-210y}^3z^2-{210y}^2z^3-105y^1z^4\] \[\ {-21y}^0z^5{+28y}^6z^0{+168y}^5z^1{+420y}^4z^2+560y^3z^3+420y^2z^4+168y^1z^5+28y^0z^6+\dots, \tag{8}\]

[ii] Expand the trinomial expression \({\left[x+y+z\right]}^{-1}\ \)using Kifilideen trinomial theorem for negative power of \(-n\) based on standardized and matrix approach. Obtain the coefficients of the series using Kifilideen coefficient table. Hence or otherwise generate the series of\(\ \ \frac{35}{82}\) (Hint: take x = 2, y = \(\frac{1}{5}\) and z = \(\frac{1}{7})\)
Solution
From Kifilideen coefficient Table 4 for negative power of \(-\ 1\), the coefficients in ascending order are \(1, -1, -1, 1, 2, 1, -1, -3, -3, -1, 1, 4, 6, 4, 1, -1, -5, -10, -10, -5, -1, 1, 6, 15, 20, 15, 6, 1, -7, -21. -35, -35, {\dots}\). The expansion gives infinite series. So the expansion of\({\mathrm{\ }\left(1+y+z\right)}^{-\ 3}\) using Kifilideen trinomial theorem of negative power of \(-\ n\ \)based on standardized and matrix approach, we have:

\[{\left[x+y+z\right]}^{-1}={x^{-1}y}^0z^0-{x^{-2}y}^1z^0-{x^{-2}y}^0z^1{{+x}^{-3}y}^2z^0{{+2x}^{-3}y}^1z^1{{+x}^{-3}y}^0z^2 -{x^{-4}y}^3z^0-3{x^{-4}y}^2z^1\] \[-{{3x}^{-4}y}^1z^2-{x^{-4}y}^0z^3+{x^{-5}y}^4z^0{+4x^{-5}y}^3z^1{{+6x}^{-5}y}^2z^2{{+4x}^{-5}y}^1z^3{{+x}^{-5}y}^0z^4-{x^{-6}y}^5z^0-{{5x}^{-6}y}^4z^1\] \[{{-10x}^{-6}y}^3z^2{{-10x}^{-6}y}^2z^3{{-5x}^{-6}y}^1z^4-{x^{-6}y}^0z^5{+x^{-7}y}^6z^0+6{x^{-7}y}^5z^1{{+15x}^{-7}y}^4z^2{{+20x}^{-7}y}^3z^3{{+15x}^{-7}y}^2z^3\] \[+6{x^{-7}y}^1z^5{{+x}^{-7}y}^0z^6{{-x}^{-8}y}^7z^0-7{x^{-8}y}^6z^1{{-21x}^{-8}y}^5z^2{{-35x}^{-8}y}^4z^3{{-35x}^{-8}y}^3z^3-21{x^{-8}y}^2z^5{{-7x}^{-8}y}^1z^6\] \[\ -{x^{-8}y}^0z^7+\dots, \tag{9}\]

\[{\left[x+y+z\right]}^{-1}=x^{-1}-x^{-2}y-x^{-2}z{{+x}^{-3}y}^2{+2x}^{-3}yz{+x}^{-3}z^2-{x^{-4}y}^3-3{x^{-4}y}^2z-{3x}^{-4}yz^2-x^{-4}z^3\] \[+{x^{-5}y}^4{+4x^{-5}y}^3z{{+6x}^{-5}y}^2z^2{+4x}^{-5}yz^3{+x}^{-5}z^4-{x^{-6}y}^5-{{5x}^{-6}y}^4z{{-10x}^{-6}y}^3z^2{{-10x}^{-6}y}^2z^3{-5x}^{-6}z^4\] \[-x^{-6}z^5+6{x^{-7}y}^5z^1{{+15x}^{-7}y}^4z^2{{+20x}^{-7}y}^3z^3{{+15x}^{-7}y}^2z^3+6x^{-7}yz^5{+x}^{-7}z^6{{-x}^{-8}y}^7-7{x^{-8}y}^6z^1\] \[\ {{-21x}^{-8}y}^5z^2{{-35x}^{-8}y}^4z^3{{-35x}^{-8}y}^3z^3-21{x^{-8}y}^2z^5{-7x}^{-8}yz^6{-x}^{-8}z^7+\dots, \tag{10}\]

[b] \(\frac{35}{82}={\left[2+\frac{1}{5}+\frac{1}{7}\right]}^{-1}\)

Using the Kifilideen expansion above where x = 2, y = \(\frac{1}{5}\) and z = \(\frac{1}{7}\) , we have:

\[\frac{35}{82}={\left[2+\frac{1}{5}+\frac{1}{7}\right]}^{-1}=\frac{1}{2}-\frac{1}{4\times 5}-\frac{1}{4\times 7}+\frac{1}{8\times 25}+\frac{2}{8\times 5\times 7}+\frac{1}{8\times 49}-\frac{1}{16\times 125}-\frac{3}{16\times 25\times 7}-\frac{3}{16\times 5\times 49}\]

\[-\frac{1}{16\times 343}+\frac{1}{32\times 625}+\frac{4}{32\times 125\times 7}+\frac{6}{32\times 25\times 49}+\frac{4}{32\times 5\times 343}+\frac{1}{32\times 2401}-\frac{1}{64\times 3125}-\frac{5}{64\times 625\times 7}\]

\[-\frac{10}{64\times 125\times 49} -\frac{10}{64\times 25\times 343}-\frac{5}{64\times 5\times 2401}-\frac{1}{64\times 16807}+\frac{1}{128\times 15625}+\frac{6}{128\times 3125\times 7}+\frac{15}{128\times 625\times 49}\]

\[+\frac{20}{128\times 125\times 343}+\frac{15}{128\times 25\times 2401}+\frac{6}{128\times 5\times 16807}+\frac{1}{128\times 117649}-\frac{1}{256\times 78125}-\frac{7}{256\times 15625\times 7}\]

\[\ -\frac{21}{256\times 3125\times 49}-\frac{35}{256\times 625\times 343}-\frac{35}{256\times 125\times 2401}-\frac{21}{256\times 25\times 16807} -\frac{7}{256\times 5\times 117649}-\frac{1}{256\times 823543}\dots, \tag{11}\]

The evaluation of the above series gives 0.426829 to 6 decimal places. Also, using calculator \(\frac{35}{82}\) gives 0.426829 to 6 decimal places. This indicates that the negative power of \(-1\) of Kifilideen trinomial theorem and coefficient of negative power of \(-1\) from the Kifilideen Coefficient table are valid.

[iii] Expand the trinomial expression \({\left[x+y+z\right]}^3\ \)using Kifilideen trinomial theorem for positive power of\(\ n\) based on standardized and matrix approach. Obtain the coefficients of the series using Kifilideen coefficient table. Hence or otherwise generate the series of\(\ \ {\left[3.74\right]}^3\) and evaluate its value (Hint: take x = 3, y = \(0.7\ \)or \(\frac{7}{10}\) and z = \(0.04\) or \(\frac{4}{100}).\)

Solution

[a] From Kifilideen coefficient Table 3 for positive power of \(3\), the coefficients in ascending order are 1, 3, 3, 1, 3, 6, 3, 3, 3, 1. So the expansion of \({\left[x+y+z\right]}^3\ \)using Kifilideen trinomial theorem based on standardized and matrix approach, we have: \[{\left[x+y+z\right]}^3=\ x^3y^0z^0{+3x}^2y^1z^0{+3x}^1y^2z^0+x^0y^3z^0{+3x}^2y^0z^1{+6x}^1y^1z^1{+3x}^0y^2z^1{+3x}^1y^0z^2{+3x}^0y^1z^2\] \[\ +x^0y^0z^3, \tag{12}\] \[\ {\left[x+y+z\right]}^3=\ x^3{+3x}^2y+3xy^2+y^3{+3x}^2z+6xyz+3y^2z+3xz^2+3yz^2+z^3, \tag{13}\] [b] \({\left[3.74\right]}^3={\left[3+\frac{7}{10}+\frac{4}{100}\right]}^3\)

Using the kif expansion above where x = 3, y = \(\frac{7}{10}\) and z = \(\frac{4}{100}\) , we have \[\ {\left[3.74\right]}^3={\left[3+\frac{7}{10}+\frac{4}{100}\right]}^3=27+\frac{189}{10}+\frac{441}{100}+\frac{343}{1000}+\frac{108}{100}+\frac{504}{1000}+\frac{588}{10000}+\frac{144}{10000}+\frac{336}{100000}+\frac{64}{1000000}, \tag{14}\]

\[\ {\left[3.74\right]}^3=27+18.9+4.41+0.343+1.08+0.504+0.0588+0.0144+0.00336+0.000064, \tag{15}\]

\[\ {\left[3.74\right]}^3=52.313624, \tag{16}\]

From calculator, the value of\(\ {\left[3.74\right]}^{\ 3}\)is also \(52.313624.\) This indicates that the positive power of \(3\) of Kifilideen trinomial theorem and coefficient of positive power of 3 of trinomial expression from the Kifilideen Coefficient table are valid.

[iv] Determine the power combination of the Kifilideen expansion of negative power of \(-3\) in the row \(13\) and column \(10\) of the Kif matrix of the Kifilideen expansion \({[x+y+z]}^{-3}.\) Hence or otherwise determine the term the generate the power combination and using Kifilideen coefficient table of negative power of \(-n\) of trinomial theorem determine the coefficient of the power combination.

Solution

[a] From the question, \(n=-3,\ r=13\ and\ c=10\)

Using Kifilideen General Row Column Matrix formula for negative power of \(\ -\ n\) [12],

\[\ CP_{rc}=n00-110\left(r-1\right)+9(r-c), \tag{17}\] \[\ {CP}_{rc}=-300-110\left(13-1\right)+9(13-10), \tag{18}\] \[\ {CP}_{rc}=-1593, \tag{19}\] [b] From Kifilideen General Term formula for negative power of \(-n\) [12],

\[t=\frac{{[n-k]}^2+\left[n-k\right]+2f+2}{2}\mathrm{\ }\]

Where,

\(t-\) the required t\({}^{th}\) term

\(k-\) the first component part of the power combination

\(i-\ \)the second component part of the power combination

\(f-\) the third component part of the power combination

\(n-\)the negative power of – n of the trinomial expression

From, \({CP}_{rc}=-1593\), \(k=-15,\ i=9\ and\ f=3\)

\[t=\frac{{[-3–15]}^2+\left[-3–15\right]+2\times 3+2}{2}\]

\(t=82\)\({}^{th\ }\)term

Using the Kifilideen Coefficient table of negative power of \(-3\),

The coefficient of \({CP}_{rc}=-1593\) of term \(82\)\({}^{th\ }\)term is \(+\ 20020\).

3.2. Demonstration on the implementation of Kifilideen theorem of matrix transformation of positive power of n of trinomial expression

Given that a term in the Kifilideen expansion of \({\left[\frac{y^3}{{{4x}^2z}^{\ 5}}-\frac{{x^7z}^{\ 4}}{2y}+\frac{e{y^6z}^{\ 8}}{x}\right]}^{\ n}\)is \(\frac{-1575}{2048}x^{\ 26}{y^{13}z}^{\ 8}\) where \(e\) is a constant value. Using Kifilideen matrix transformation method of positive power of \(n\) of trinomial expression. Find

[i] the power combination

[ii] the degree of the positive power of the trinomial expression

[iii] the t\({}^{th}\) term

[iv] the value of e.

Solution

[i] Trinomial expression: \({\left[4^{-1}x^{-2}y^3z^{-5}-{2^{-1}x}^7y^{-1}z^4+ex^{-1}y^6z^8\right]}^n\)

Power combination to be obtained: \(\ \ \ \ \ \ k\ \ \ \ i\ \ \ \ \ \ f\)

\(t\)\({}^{th}\) term of the power combination: \(\frac{-1575}{2048}x^{\ 26}{y^{13}z}^{\ 8}\)

Using the kifilideen matrix transformation method, so

\[ \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]: \left[ \begin{array}{ccc} -2 & 7 & -1 \\ 3 & -1 & 6 \\ -5 & 4 & 8 \end{array} \right] \left[ \begin{array}{c} k \\ i \\ f \end{array} \right]=\left[ \begin{array}{c} 26 \\ 13 \\ 8 \end{array} \right], \tag{20}\]

Also, \(k+i+f=n\)

Using Crammer’s rule, so \[k=\frac{\Delta k}{\Delta }=\frac{\left| \begin{array}{ccc} 26 & 7 & -1 \\ 13 & -1 & 6 \\ 8 & 4 & 8 \end{array} \right|}{\left| \begin{array}{ccc} -2 & 7 & -1 \\ 3 & -1 & 6 \\ -5 & 4 & 8 \end{array} \right|}\]

\[\ k=\frac{-1284}{-321}, \tag{21}\]

\[\ k=4, \tag{22}\]

\[i=\frac{\Delta i}{\Delta }=\frac{\left| \begin{array}{ccc} -2 & 26 & -1 \\ 3 & 13 & 6 \\ -5 & 8 & 8 \end{array} \right|}{\left| \begin{array}{ccc} -2 & 7 & -1 \\ 3 & -1 & 6 \\ -5 & 4 & 8 \end{array} \right|}\]

\[\ i=\frac{-1605}{-321}, \tag{23}\]

\[\ i=5, \tag{24}\]

\[f=\frac{\Delta f}{\Delta }=\frac{\left| \begin{array}{ccc} -2 & 7 & 26 \\ 3 & -1 & 13 \\ -5 & 4 & 8 \end{array} \right|}{\left| \begin{array}{ccc} -2 & 7 & -1 \\ 3 & -1 & 6 \\ -5 & 4 & 8 \end{array} \right|}\]

\[\ f=\frac{-321}{-321}, \tag{25}\]

\[\ f=1, \tag{26}\] So, the power combination \(=kif=451\)

[ii] the positive power of n of the trinomial expression \(=n\)= \(k+i+f=4+5+1\) \[ \n=10, \tag{27}\]

[iii] \(C_p\) \(=kif=451\) \[\ k=4,\ i=5\ and\ f=1, \tag{28}\] \[\ n=k+i+f=4+5+1=10, \tag{29}\]

From Kifilideen general term formula [19],

\[t=\frac{-f^2+2fn+3f+2i+2}{2}\]

Where,

\(C_p-\ \) the given power combination

\(k,i,f-\) the component parts of the power combination

\(f-\)the third component part of the given power combination

\(i-\)the second component part of the given power combination

\(n-\)the degree of the positive power of the trinomial expression

\(t-\ \)the term of the given power combination to be determined

\[t=\frac{-{[1]}^2+2\times 1\times 10+3\times 1+2\times 5+2}{2}\]

\(\ \ \ \ \ t=17\)\({}^{th}\) term

OR

Using Kifilideen general power combination formula [11,20,21],

\[\ C_p =kif=451, \tag{30}\] \[\ k=4,\ i=5\ and\ f=1, \tag{31}\] \[\ n=k+i+f=4+5+1=10, \tag{32}\] \[\ a=f=1, \tag{33}\] \[\ m=\frac{a}{2}\left[2n-a-1\right]=\frac{1}{2}\left[2\times 10-1-1\right]=9, \tag{34}\] \[C_p=kif=-90t+ 81 a+ 90m +n90\]

Where,

\(C_P\) \(\mathrm{-}\) Power combination, \(kif\)

t- nth term of the kifilideen trinomial theorem

a- the power of the third digit, \(f\) of the power combination or the value of the third digit of the column or group the term fall into

a and m – are constant values for a particular group or column of the matrix

n- the power of the trinomial expression

\(k,\ i\ \)and \(f\) \(–\) the first, second and third component part of the power combination

\(451\) \(=kif=\) \(90t + 81\) \(\times 1\) + 90 \(\times 9\) \(+\) 1090

\(t=17\)\({}^{th}\) term

OR

Using alternate Kifilideen general power combination formula [19],

From the question, \(C_p\) \(=kif=451\)

\(k=4,\ i=5,\ f=1\) and \(n=10\)

\({\ C}_p\) \(=kif=\) \(90t\) \(-\) 45 \(f^2\) \(+36f+\) 90 \(fn+\) \(n90\)

\(451=\) \(\mathrm{-}\) \(90t\) \(-\) 45 \({[1]}^2\) \(+36\times 1+\) 90 \(\times 1\times 10+\) 1090

\(t=17\) \({}^{th}\) term

[iv] From the question, \(u\ =\ 4,\ v\ =\ 1,\ w=e\ and\ q=\frac{-1575}{2048}\)

Using Kifilideen matrix transformation method,

\[\ {}^n_{kif}{Cu^k}v^iw^f=q, \tag{35}\] \[\ {}^{10}_{451}{C{[4^{-1}]}^4}{[-2^{-1}]}^5{[e]}^1=\frac{-1575}{2048}, \tag{36}\]

From the Kifilideen coefficient table, the coefficient of the 17\({}^{th}\)\({}^{\ }\)term on the n = 10 column is 1260

\[\ 1260\boldsymbol{\times }\frac{1}{256}\times -\frac{1}{32}\boldsymbol{\times }e=\frac{-1575}{2048}, \tag{37}\]

\[\ e=5, \tag{38}\]

4. Conclusion

Kifilideen theorem of matrix transformation of positive power of \(n\) of trinomial expression in which three variables \(x,\ y\) and \(z\) are found in parts of the trinomial expression was established. The research work established Kifilideen coefficient tables for positive and negative powers of \(n\) and\(-n\) of Kifilideen trinomial theorem base on standardized and matrix approach. The development of theorem of matrix transformation of positive power of \(n\) of trinomial expression would make the process of evaluating such positive power of \(n\) of the trinomial expression easy. The inaugurated tables had been fully utilized to generate series of expansion for positive and negative power of binomial and trinomial expressions. The Kifilideen coefficient tables are handy and effective in generating the coefficients of terms and series of the Kifilideen expansion of trinomial expression of powers of \(n\) and \(-n\).