Binomial Theorem

The coefficient of x^n-ry^r is given by

           _nC_r  =       n!
                      (n-r)!r!
 
 

Pascal's Triangle

                                     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

Each row is obtained by adding consecutive numbers in the row above it.
 

Example:   Expand  (x – 4)⁵

      x⁵ + 5x⁴(-4) +  10x³(-4)² + 10x²(-4)³ + 5x(-4)⁴ + (-4)⁵

 =  x⁵ -  20x⁴  +  160x³  -  640x²  +  1280x  -  1024
 

Note:  One way to get the coefficients of the first line is to take each coefficient times the first power in that term divided by one more than the second exponent in that term and make it the coefficient of the next term.
                                              ¥
Binomial Series:   (a + b)ⁿ  = å (_nC_k ) a^n-kb^{k
                                                            k=0}

An Unusual Example:
(1 + x)^-4 =  _-4 C₀(1)^-4-0x⁰  +  _-4 C₁(1)^-4-1x¹ + _-4 C₂(1)^-4-2x² +  _-4 C₃(1)^-4-3x³ +  ...

 _-4 C₀  =    (-4)!       =  -4 (-4-1) (-4-2) (-4-3)...  =  -4 (-5) (-6) (-7)...  =  1
               0! (-4)!         -4 (-4-1) (-4-2) (-4-3)...       -4 (-5) (-6) (-7)...

_-4 C₁  =    (-4)!       =  -4 (-5) (-6) (-7)...  =  -4
               1! (-5)!          (-5) (-6) (-7)...

_-4 C₂  =    (-4)!       =  -4 (-5) (-6) (-7)...  =  10
               2! (-6)!         2 (-6) (-7)...

_-4 C₃  =    (-4)!       =  -4 (-5) (-6) (-7)...  =  20
               3! (-7)!         6 (-7)...

so the series becomes   (1 + x)^-4 =  1 - 4x + 10x² - 20x³ +...
 
 

Another unusual example is (1 + x)^2/3

It works the same as the one above.  (2/3)! = (2/3)(-4/3)(-7/3)....

The answer is:  1 + (2/3)x - (1/9)x² + (4/81)x³ + ...  See if you can arrive at this answer.