December 19th, 2006, 02:41 PM  #1 
Member Joined: Nov 2006 From: BIG APPLE Posts: 63 Thanks: 0  terms
find the third term of (2x  3y^2)^5

December 19th, 2006, 03:01 PM  #2 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
The third term will be 5C3a^3b^2 = 10*(2x)^3*(3y)^2 =720x^3y^2 If you don't understand why, feel free to ask, but be specific to give us the best chance of giving you useful information. 
December 19th, 2006, 03:43 PM  #3  
Member Joined: Nov 2006 From: BIG APPLE Posts: 63 Thanks: 0  Quote:
how did you get 5c3a^3b^2 and arrive at that answer?  
December 19th, 2006, 04:18 PM  #4 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
5C3 refers to the "combination function." nCr=n!/[r!(nr)!] where n! (pronounced "n factorial")=n(n1)(n2)(n3)...*3*2*1 for all natural numbers n, and 0!=1. In (a+b)^n, the k^th term is always nCk[a^(nk)b^k]. Note that nCr is undefined if r>n, because (1)! makes no sense.

December 19th, 2006, 04:23 PM  #5 
Member Joined: Nov 2006 From: BIG APPLE Posts: 63 Thanks: 0 
oh i see thank you so very much


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