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Bilal
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#1
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#1
Could i please get help on the following question; don't have a clue how to do it...

Use the binomial Theroem to find the middle term in the expansion of:

(2x - 1/x)^10

Thanks for any help.
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JamesF
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There are 11 terms, the middle term is the 6th, remmber that the first term has binomial coefficient nC0, not nC1, so the 6th term has binomial coefficient nC5.
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Bilal
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(Original post by JamesF)
There are 11 terms, the middle term is the 6th, remmber that the first term has binomial coefficient nC0, not nC1, so the 6th term has binomial coefficient nC5.
Could you elaborate please? :confused:

Do I use pascals triangle...?
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BLUREMI
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(Original post by Bilal786)
Could you elaborate please? :confused:

Do I use pascals triangle...?
i think u have to use the formula
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Bilal
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Anyone? :confused:

I've tried but not getting anywhere with this.
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J.F.N
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(r+1)^th term = nCr.(2x)^(n-r).(-1/x)^r

You want the 6th term. So r=5
So its 10C5.(2x)^5.(-1/x)^5=252.32.x^5.-1/x^5=-8064
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Bilal
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(Original post by J.F.N)
(r+1)^th term = nCr.(2x)^(n-r).(-1/x)^r

You want the 6th term. So r=5
So its 10C5.(2x)^5.(-1/x)^5=252.32.x^5.-1/x^5=-8064
Cheers. Thanks for the help.
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