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Coefficents

AM4R

Posted 20 January 2007 - 12:23 PM

Hi

Could someone help me with a this type of question, I dont know really know how to this as I was off school when it was done and the teaher just skimmed over it, so the more deatil the better.

Thanks

Find the coefficient of x^10 in the expansion ( 3x^2 + 4/x^3)^10

ad absurdum

Posted 20 January 2007 - 06:43 PM

(3x^2 + 4x^{-3})^{10} = \displaystyle\sum_{r=0}^{10} {10\choose r} (3x^2)^{10-r} (4x^{-3})^r

So the general term of the expansion is {10\choose r} (3x^2)^{10-r} (4x^{-3})^r. Now we need to find the value of r that gives a term in x^{10}, so what you can do is ignore everything except the x's and set this equal to x^{10}, giving:

(x^2)^{10-r}\times (x^{-3})^r = x^{10}
x^{20-2r-3r} = x^{10}

So r = 2. Use this value of r in the general term now to give:

{10\choose 2} (3x^2)^{8} (4x^{-3})^2

Evaluate this, and you can take the number in front of x^{10}, which is the coefficient.

AM4R

Posted 21 January 2007 - 09:54 PM

Thanks for that. Really helps smile.gif

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