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    Multiple Binomial Theorem

    ABTRACT
    Sometimes in life, we are faced with the challenges of permutation and combination of diverse scenarios. This led to the evolvement of the Binomial Theorem for solving equations of the form (X + Y)n. After a critical study of this theorem, it was discovered that some modifications are required to solve for multiple binomials of the form (X + V1)(X + V2)(X + V3)…(X + Vn). In this project, a valid formula for dealing with these types of binomials was postulated and a step by step methodology of its application was documented. It is believed that this theorem will find wide applications in resolving some scientific challenges.

    INTRODUCTION
    Scientists of old did a lot of researches to better their lives and that of those around them. These efforts are responsible for the relative comfort we enjoy today. From the few researches I have done which are most times triggered by challenges before me, I discovered that what are yet to be known are more than all presently known. Who will discover them? Formula to directly expand multiple binomial was one of them.   

    OMOROGHOMWAN MULTIPLE BINOMIAL THEOREM
    A MULTIPLE BINOMIAL OF THE FORM (X + V1)(X + V2)(X + V3)…(X + Vn) CAN BE EXPANDED DIRECTLY WITHOUT FIRST OPENING BRACKETS BY SYSYEMATICALLY COMBINING THE CONCERNED VARIABLES IN THEIR NORMAL POWER EQUIVALENT AND ORDERLY ADDED IN THE NUMBER OF TIMES OF THEIR CORRESPONDING BINOMIAL COEFICIENTS AND OPERATING THE CONSTANT AS USUAL.

    i.e.       (X + V1)(X + V2)(X + V3)…(X + Vn) = r=0n { Xn-r    (nCr li Vr   )} -     (1)
                                                                                                                                         

     nCr:  This is the number of times the variables (V) should be added when combined in r’s which is the normal binomial coefficient
    X:     The constant
    V:     The variable eg. a ,b, c.
     i:      First variable(V1  ) which is a in our case
     l:      Last variable (Vn)                                 
     n:     no of brackets in a given case
     r:      The number of the variable that should be combined as an entity e.g. when r is 2 then the combination should look like this ab + bc+ cd+…….and when it is 3, the combination should be like this abc + bcd+ cde +…………

    ANALYSIS
    SIMPLE BINOMIAL is an expression in which the sum of two constants are raised to a given power e.g.
                                                                                                                 
    (y+q)2, (t-u)3,(s+r)21,

    and it has a formula eg.

                           
    (a+b)n=        r=0n nCr an-r br                   -              -              -                      (2)
                           

    A MULTIPLE BINOMIAL is the product of more than one bracket which carries the sum of a constant and variables e.g.

    (x+a)(x-b)(x+c) -              -              -              -              (3)

    Here x is the constant while a,b,and c are the variables (V’s)


    SYSTEMATIC COMBINATION

    DIRECT EXPANSION is a process by which a simple or multiple binomial expression is expanded in which the constant to a given degree carries the sum of all its coefficients in one place e.g.
          
    i.                  (x+a)3 =x3 + x2(3q) + x1(3q2) + x0(q3)          -              -              -              (4)

    ii.                (x + a) (x + b) (x + c)= x3 + x2(a + b + c) + x1(ab + ac+bc) + x0(abc)  -       -       (5)

    We assumed x to be the constant in equation 4 above and also assuming q to be variables while in equation 5 we also assume x is the constant but a, b and c are the variables. . In equation 4 above, x2 is a constant to a given power (degree 2) with the sum of coefficient (q + q + q =3q) while in equation 5, x2   was a constant to degree 2 with the sum of its coefficients (a + b + c).

    NORMAL POWER EQUIVALENT: Consider equation 6

     (x+q)4  -             -              -              -              -              -              (6)

                                       
    Recall    (a+b)n=        r=0n nCr an-r br
                                      
    Where n = 4
                                   

    = x+  x3(4q)   +  x2(6 q2) + x(4q3) +  q4

    the sum of x2 coefficients is  6q2 (ie q2+ q2 + q2+ q2+ q2+ q2)

    also consider equation 7 below

    (x + a) (x + b) (x + c) (x + d)          -              -              -              -              (7)

    The sum of x2 coefficients are (ab + bc + cd  + da + ac +bd). Refer to equations 9 and 17.

    This is true because if

    q = a = b =c = d

    then

    ab + bc + cd + da + ac +bd

    = a2+ a2+ a2+ a2+ a2+ a2   (6a2)

    = b2+ b2+ b2+ b2+ b2+ b2   (6b2)

    = c2+ c2+ c2+ c2+ c2+ c2   (6c2)

    =d2+ d2+ d2+ d2+ d2+ d2   (6d2)  

    =   q2+ q2+ q2+ q2+ q2+ q2   (6q2)


    We can then have a relationship like this


    (x+q)4 = (x+a) (x + b) (x + c) (x + d)

    This time we focus on the variables. The truth is that degree 2 in a simple binomial will combine in twos in the corresponding multiple binomial likewise degree 3 will combine in threes and so on, that is the normal power equivalent.


    ORDERLY ADDED ……Some binomials (simple and multiple) may contain negative variables. The additional of these special types have to be done orderly .When negative variables combine in odd numbers of time, it take   –   and takes   +   when combining in even numbers of ways or power as seen in equation 8 and 9

    i.                     (x-q)4     -              -              -              -              -              (8)

                             = x4    -   x3 (4q) + x2(6q2)    -     x(4q3) + q4

    ii.                   (x-a)  (x-b)  (x-c)  (x-d)   -              -              -              -              (9)   

     = x-x3(a + b + c + d) + x2(ab+bc+cd+da+ac+bd) - x (abc + bcd + cda + dab) + abcd.

     Note the relationship between equations 8 and 9.

    IN THE NUMBER OF TIMES OF THEIR CORRESPONDING SIMPLE BINOMIAL COEFFICIENTS: Look again at equation 8 and 9 above, they have something in common which is x, the constant.  Let take x4 in i to be the same x4 in equation 9 likewise x3, x2,x, now the simple binomial  coefficient in equation 8 were 1, 4, 6, 4, 1 respectively alongside with the variable q to various degrees but the case in equation 9 is more complicated, therefore it has to be systematically combined and added in the number of times of its equivalent simple binomial coefficients. This was why the coefficient of x3  in equation 8 changed from 4q (q+ q+ q +q ) to a + b + c + d in equation 9 correspondingly x2 ‘s  changed from 6q     (ie q2+ q2+ q2+ q2+ q2+ q2) in i to  ab + bc + cd  + da + ac +bd)  in equation 9 just as x’s changed from 4q3   (ie q3+ q3+ q3+ q3+ q3+ q3) in equation 8 to abc + bcd + cda + dab) in equation 9.

    It therefore follows that

    equation 8 = equation 9

    IF q = a = b = d.

    OPERATING THE CONSTANT AS USUAL means that the constant follows normally whether in simple or multiple binomial operations. This has already been dealt with as seen above

    OMOROGHOWMAN’S RULES FOR MULTIPLE BINOMIAL SYSTEMATIC COMBINATION
    1.       In each bracket, each variable must appear in the same number of times as the rest. This is equal to the coefficient number directly above the bracket in the normal binomial form when using the Omoroghomwan’s triangle as shown in fig. 2.
           Omoroghomwan’s Triangle is a modification of Pascal’s Triangle of Fig 1. below.


    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

    Fig. 1 Pascal’s Triangle


          1
                                                                                         1         1
                                                  *                         1          2         1
    β               1         3          3         1
                                                      1          4         6         4          1
                                            1        5        10        10        5         1
                             1             6       15       20       15        6          1

    Fig. 2. Omoroghomwan’s Triangle


    See equations 10 and 11 for explanation.

    Eg (x-q)3               -              -              -              -              -              (10)

                               =                                           x3 _ x2 (3q) +  x(3q3) –  q3  

    Binomial coefficient (      β       )  =        1            3            3          1


     (x-a) (x-b) (x-c)                      -              -              -              -              (11)

                                 =                                       x3 - x2(a - b - c) + x (ab + ab + ca) –abc

    The number of each variable appearances

    In each bracket    (*)                 =                              1                      2                   1

    Where    *   stands for the number of times each variable should /must appear in the bracket as predicted by Omoroghomwan following his triangle .                              
                                                                                                                
    While          β              represents the normal binomial coefficients as derived from Pascal's Triangle. 


    2.       There must not be more than one identical combination 
                   eg ace  =  eca  =  eac  =  aec.     Only one must appear.

    3.       One variable must not appear more than once in each combination eg adba is wrong just as apqbq  is

    4.       For easy and accurate combination note the following steps;


    1st , start by combining the variables alphabetically until the last become the head e.g. combine abcdef in fours as  follows abcd ,bcde, cdef,defa, efab, fabc, but when this has ended then try the next step.

    2nd, combine alphabetically but skipping the next alphabet to the head until the cycle is completed e.g.  acde, bdef,cefa,dfab, eabc,fbcd, when this try to skip 2, 3, 4, and so on until you discover that such no longer holds.

    3rd, combine by keeping the first two variables constant before skipping the third alphabet till the cycle is completed e.g. abcde, bcdef, cdefa. If there are still more try 2, 3, alphabets and so on until it no longer holds. Then start again by keeping the first 3, 4 alphabets and so on using the above manipulation i.e. the skipping of 1, 2, 3 and so on.

    4th, combine alphabetically skipping one after each variable e.g. In combining abcdef in threes, we have something like abcde   and   bcdef. If there are still more then keep the first 2, 3 alphabets constant before
    Skipping one, two, three…… n alphabets.

    5th, stop the combination if the numbers of combined entities are equal to the numerical coefficient in the corresponding normal binomial using the Pascal’s Triangle.


      
    WORKED EXAMPLES

    1.       i.      (x +q)        -              -              -              -              -              (12)

                              
    i.e    (a+b)n    =    r=0n nCr an-r br
                                
                  Where n = 2

                                         =   x2+ (2q) + q2
                 β     =             1       2       1

    ii.        (x+a) (x+b)              -              -              -              (13)



    i.e.       (X + V1)(X + V2) = r=0n { X2-r   (2Cr ba Vr )}


                                       =  x2 + x(a + b)   + ab

                      *               =               1            1



    2.       i.      (x + q)       -              -              -              -              (14)

                               
    i.e    (a+b)n=        r=0n nCr an-r br
                                
                Where n = 3


                                    =   x+  x2(3q) +(3q2) +  q3
                   β          =  1           3           3          1


    ii.      (x + a) (x + b) (x + c)               -              -              -              (15)

                                                         
    i.e.       (X + V1)(X + V2)(X + V3)= r=0n {X3-r   (3Cr ca Vr)}
                                                                                                                        

                                  = x3 + x2 (a + b+c)  + x (ab + bc + ca)+ abc
        *                   =                        1                      2                    1

    3.       i.      (x-q)4                     -              -              -              -              (16)


                               
    i.e    (a+b)n=        r=0n nCr an-r br
                               
                  Where n = 4
                                   

                                         = x-  x3(4q)   +  x2(6 q2) + x(4q3) +  q4
                   β            =                 1           4                6              4           1




    ii.             (x-a) (x-b) (x-c) (x-d)            -              -              -              (17)  

                                                                      
    i.e.       (X + V1)(X + V2)(X + V3)(X + V4) = r=0n {X4-r   (4Cr da Vr )}
                                                                                                                                 


      = x4 -  x3 (a +b+c+d)  + x2 (ab+bc+ca +da+ ac+bd) –x( abc + bcd + cda + dab) +abcd

    *    =                 1                                        3                                 3                                     1


    4.        i.     (x-q)5                     -              -              -              (18)

                               
    i.e    (a+b)n=        r=0n nCr an-r br
                                  
                 Where n = 5


      = x5  - x4(5q) +  x3(10q2)   +  x2(10q3) + x(5q4) -  q4

                                                                             
     ii.            (x-a) (x-b) (x-c) (x-d) (x-e)    -       -            -              -              (19)

                                                                                      
    i.e.       (X + V1)(X + V2)(X + V3)(X + V4)(X + V5) = r=0n {X5-r   (5Cr ea Vr )}
                                                                                                                                                 
                                                                                                                                        

         = x6 - x4 (a + b + c + d + e) + x3 (ab + bc + cd + de + ea + ac+ bd + ce + da + eb)

      
           -x2( abc + bcd + cde + dea + eab+ acd + bde + ced + dab + ebc)


                + x(abcd +bcde +cdea + deab + eabc ) – abcde


    5.                    i.             (x-q)                   -              -              -              (20)


                                
    i.e    (a+b)n=      r=0n nCr an-r br
                                 
                      Where n = 6


                            = x6 - x5(6q) + x4(15q2)   +  x3(20q3) + x2(15q4) - x(6q5) +  q6

                                                                               
     ii.            (x-a) (x-b) (x-c) (x-d) (x-e) (x-f)                                 -              -              (21)


                                                                                                 
    i.e.       (X + V1)(X + V2)(X + V3)(X + V4)(X + V5)(X + V6) = r=0n { X6-r   (6Cr fa Vr )}
                                                                                                                                                             

                                                  
                = x6 - x5 (a + b + c + d + e + f) + x4 (ab + bc + cd + de + ef + fa + ac + bd + ce

                   + df + ea+ fb + ad + be + cf) -x3( abc + bcd + cde + def + efa + fab + acd

                   + bde + cef + dfa + eab + fbc + ade + bef + cfa + dab + ebc + fcd + ace +bdf )

                   + x2 (abcd +bcde +cdef + defa + efab + fabc + acde + bdef + cefa + dfab

                    + eabc + fbcd + abde +bcef + cdfa ) –x( abcde + bcdef +cdefa + defab

                     + efabc +fabcd ) +abcdef

    Recall that;
    *        The number of times each variable should /must appear in the bracket as predicted by Omoroghomwan following his triangle.                               
                                                                                                                
      β                 The normal binomial coefficient.



    QUESTIONS

    i.                     a, b, c d & e combine ten times in 3’s but a,b,c ,d,e, & f combine  twenty times in 3’s.
                   Why?

    Answer: Because of the introduction of another variable (f)

    ii.                   What does this implies.

    Answer: It implies that the new variable (f) must combine ten times with the rest to make them twenty entities in that bracket.

    Also note that ten is Omoroghomwan’s Triangle parameter while twenty is Pascal’s.


    iii.                  Assume x=10, a=1, b=2, c=3, d=4
    Evaluate
    (x-a)(x-b)(x-c)(x-d)         -              -              -              -              (22)

    Solution:
    Substituting values, equation 22 becomes
    (10-1)(10-2)(10-3)(10-4)
    =9x8x7x6
    =3024

    Also from worked example 3, equation 22 becomes,

    X4-x3(a+b+c+d)+x2(ab+bc+cd+da+ac+bd)-x(abc+bcd+cda+dab)-abcd     -    -     (23)

    Substituting values into equation 23, we have.

    104-103(1+2+3+4)+102(1x2+2x3+3x4+4x1+1x3+2x4)-10(1x2x3+2x3x4+3x4+1+4x1x2)+(1x2x3x4)

    =104-103x10+102x35-10x50+24

    =104-104+3500-500+24

    =3024

    Therefore, equation 22 = equation 23


    CONCLUSION
    It is hoped that this theorem will find wide usage in the fields of engineering and medicine.  



    REFERENCE
    The Mathematical Database; Binomial Theorem.                 

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