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Math Help - [SOLVED] Question regarging Binomial Theorem

  1. #1
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    [SOLVED] Question regarging Binomial Theorem

    Find n so that the coefficients of the eleventh and thirteenth terms in the expansion of (1+x)^n are the same.

    I need some help regarding this question, any help would be greatly appreciated.
    Thx in advance!
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  2. #2
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    Quote Originally Posted by Solid8Snake View Post
    Find n so that the coefficients of the eleventh and thirteenth terms in the expansion of (1+x)^n are the same.

    I need some help regarding this question, any help would be greatly appreciated.
    Thx in advance!
    the expansion will have 23 terms.

    what value of n will yield 23 terms?

    hint : look for a pattern in the symmetry and the number of coefficients in the rows of Pascal's triangle ...
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  3. #3
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    Hello, Solid8Snake!

    Another approach . . .


    Find n so that the coefficients of the 11^{th} and 13^{th} terms in the expansion of (1+x)^n are equal.
    The coefficients of the 11th and 13th terms are: . _nC_{10}\,\text { and }\,_nC_{12}

    We have: . _nC_{10} \:=\:_nC_{12} \quad\Rightarrow\quad \frac{n!}{10!(n-10)!} \;=\;\frac{n!}{12!(n-12)!} \quad\Rightarrow\quad 10!(n-10)! \;=\;12!(n-12)!

    Divide by 10!\!:\quad (n-10)! \;=\;12\cdot11\cdot(n-12)!

    Divide by (n-12)!\!:\quad (n-10)(n-11) \;=\;12\cdot11 \quad\Rightarrow\quad n^2 - 21n - 22 \;=\;0

    . . (n+1)(n-22) \;=\;0 \quad\Rightarrow\quad n \;=\;-1,22


    Therefore: . n \:=\:22

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