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December 19th, 2006, 02:41 PM   #1
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find the third term of (2x - 3y^2)^5
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December 19th, 2006, 03:01 PM   #2
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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.
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December 19th, 2006, 03:43 PM   #3
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Quote:
Originally Posted by roadnottaken
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.
yeah i dont understand at all

how did you get 5c3a^3b^2 and arrive at that answer?
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December 19th, 2006, 04:18 PM   #4
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5C3 refers to the "combination function." nCr=n!/[r!(n-r)!] where n! (pronounced "n factorial")=n(n-1)(n-2)(n-3)...*3*2*1 for all natural numbers n, and 0!=1. In (a+b)^n, the k^th term is always nCk[a^(n-k)b^k]. Note that nCr is undefined if r>n, because (-1)! makes no sense.
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December 19th, 2006, 04:23 PM   #5
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oh i see thank you so very much
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