Binomial Theorem

`    Factorial Notation  :For a natural number n , the factorial 

   of n is written as n! or ∟n & read as " n factorial " is the 

   productof n natural numbers from 1 to n .

            i.e. n! =n (n-1)(n-2)(n-3)(n-4) ........4 . 3 . 2 . 1

                       =n (n-1)!

                       =n.(n-1).(n-2)!

          e.g.

               0!=1

               1!=1 

               2!=2✕1=2.1=2

               3!=3✕2✕1=3.2.1=6

               4!=4✕3✕2✕1=4.3.2.1=24

               5! =5✕4✕3✕2✕1  =5.4.3.2.1=120

               6!=6✕5✕4✕3✕2✕1=6.5.4.3.2.1=720

               7! =7✕6✕5✕4✕3✕2✕1=7.6.5.4.3.2.1=5040

                        ......... so on.                    

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COMBINATION : A Combination of a set of n object taken r 

        at a time without repetition is known as Combination in                        Mathematics

        i.e.   Out of n object if r object is selected at time , 

           where order is not important. It is denoted by 

            nCr or C(n ,r) or nCr .

                        nCr =        n!        =    n!/r!(n-r)!

                                   r! (n-r)!

                e.g.    10C4 =10!/ 4! (10-4)! 

                                    =    10.9.8.7.6!    

                                           4.3.2.1.  6!

                                             =  10.9.8.7       = 210

                                           4.3.2.1

           The  Above calculation is according to the defination

             of Combination.

             The Calculation can also be done like this .

                  10C4 =    10.9.8.7        

                             4.3.2.1 

                20C6 =  20.19.18.17.16.15   

                               6.5.4.3.2.1

  i.e. If the value of r is 4 then just write 

Product of n to (n-3) narural number in the numerator 

and Product of 4 natural number from1to 4 in denominator 

     as in 1st example.

                 Note : nC0=nCn=1 & nC1=n

                 Bionomial Theorem  :

                             (a+b)¹ = a + b

                             (a+b)²=  a² +2ab+ b²    

                   (a+b)³= a³ + 3a²b + 3ab² + b³

                   (a+b)⁴= ?

           When the higher powers of  (a+b) comes then it 

       becomes difficult for us to remember. If someone 

           asks us what is 29th power of (a+b) =?

     Then it becomes difficult for us to answer . 

     At that time we can use Binomial Theorem

        to Find Higher Powers of (a+b) or any variable.

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     Binomial Theorem : For a,b∊R and n∊N

(a+b)ⁿ =nC0aⁿbº +nC1aⁿ⁻¹b¹ +nC2aⁿ⁻²b²+nC3aⁿ⁻³b³ +........+nCnaºbⁿ

  For Example :

 (a+b)⁴ = 4C0 a⁴ b⁰+4C1a³b¹ +4C2 a²b² +4C3 a¹b³+4C4 a⁰b⁴

            = a⁴ b⁰+4C1a³b¹+4C2 a²b² +4C3 a¹b³ +b⁴    (4C0=4C1=1)

(a+b)⁴= a⁴ +4a³b¹ +6a²b² +4a¹b³ +b⁴   

  Simillarly we can calculate          

 1)    (a+b)⁵ 

= a⁵+5a⁴b¹ +10 a³b² +10a²b³ +5a¹b⁴+b⁵ 

2) (a+b)⁶ 

 =a⁶+6a⁵b¹+15a⁴b²+20a³b³+15a²b⁴+6a¹b⁵+b⁶

3) (a+b)⁷

=a⁷+7a⁶b¹+21a⁵b²+35a⁴b³+35a³b⁴+21a²b⁵+7a¹b⁶+b⁷

 This way we can calculate any power of (a+b) by using Binomial Expansion.

     Note :

       1) In the expansion of (a+b)ⁿ contains (n+1) terms .

   For Example (a+b)² contain 3 terms , (a+b)³ contains 4 terms & so on .

    Expansion contain one more term  than the power of expansion.

        2) First term is alway aⁿ & last term is bⁿ  .

            For Example in the expansion of  (a+b)² first term is a² 

            and last term is b².

     In Example of (a+b)³ , First term is a³ and last term is b³ so on .

      3) In each term , the sum of  power of each term

            i.e a &b is alway n

     For example in the expansion of (a+b)², power of 1st term is 2 , 

      in second term sum of power of a & b is 2 and 

      Power of last term is also 2.

      simillarly in (A+b)³& so on

       4) In the successive term in the expansion we can

    see that power of a decreases by 1 and Power of b increases by one.

Binomial Theorem can used for negative term also .

          For example 

    (a-b)⁴= a⁴-4a³b¹+6 a²b² - 4a¹b³ +b⁴   

   (a-b)⁵= a⁵-5a⁴b¹+10 a³b²-10a²b³ +5a¹b⁴- b⁵ 

    Here we can note that First term is +ve , 

     second term is negative & so on. i.e. 

        Alternate terms are +ve & -ve .

   Binomial can also used for fractional & negative powers also .

        1/1+x = (1+x)⁻¹  =1-x+x²-x³+x⁴-x⁵+.......

        1/1-x =(1-x)⁻¹  = 1+x+x²+x³+x⁴+x⁵+.......

  1/(1+x)²  = (1+x)⁻²  =1-2x+3x²-4x³+........

  1/(1-x)²  =(1-x)⁻²  =1+2x+3x²+4x³+......

1/√(1+x)  =(1+x)⁻¹/²  = 1+x/2-x²/8+x³/16+........

  1/√(1-x) =(1-x)⁻¹/² =  1-x/2-x²/8-x³/16-.........

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