Expert: Sherry Wallin - 5/23/2010

Question
(a+b)^5=C(5,0)*a^5+C(5,1)a^4 *b.......+b^2

sherry pls please decsribe the the first nd second term.

i.e C(5,0)*a^5+C(5,1)a^4 *b

Answer
Hi Pratap~
    Do you know what Pascal's Triangle is?
         1
      1     1
    1    2    1
   1  3    3    1
 1  4    6   4   1
1  5   10   10  5   1
.....................
Notice if you look at any two elements next to each other in a row that they add up to the number above them. Also you number the rows starting with zero instead of one, so the rows above are rows 0, 1, 2, 3, 4, 5 and there are six rows written out. These rows go on forever in that pattern. This is the intersting thing about Pascal's triangle: If you look at the row number then it represents the coefficinets of (a+b)^(row number). For example look at row number 2, there are the coefficients 1 2 1 for (a+b)^2. Check it out, expand (a+b)^2
= (a+b)(a+b) = a^2 + 2ab + b^2. But each row is also represented by combinations nCr where n repreents the row number so n the same row we just discussed we have 2C0  2C1  2C2 which is 1 2 1. So n your example above (a+b)^5 you are in row 5 (which is really the 6th row) and its coefficients should be 1 5 10 10 5 1 or 5C0 5C1 5C2 5C3 5C4 5C5;
5C0 = 5!/0!5! = 1; 5C1 = 5!/1!4! = 5*4!/1!4! = 5; 5C2 = 5!/2!3! = 5*4*3!/2!3!= 20/2 = 10;
5C3 = 5!/3!2! = 5*4*3!/3!2! = 20/2 = 10; 5C4 = 5!/4!1! = 5*4!/4!1! = 5; and finally
5C5 = 5!/5!0! = 1. See you get the row n coefficients!

Math Prof