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Math Help - two problems in binomial theorem

  1. #1
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    two problems in binomial theorem

    1)In the expansion of (2x-3)^1^5 if  13T3 + 10T4 + T5 = 0 where Tn is the term number n
    find the values of X that satisfy the given
    ....My solution : x= 0.5 , 4.5 , 0
    Guide answer : 0.5 , 4.5 only ???


    2)If the ratio between the two middle terms in the expansion of (1+x)^2^7 is 1:4 find the value of X
    My solution : x=4 or 0.25
    Guide answer : 4 only ???
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  2. #2
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    Re: two problems in binomial theorem

    Hey mido22.

    Can you show your working for the problems?
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  3. #3
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    Re: two problems in binomial theorem

    1) t3=945 (2x)^1^3    ,   t4=-12285(2x)^1^2   ,   t5=110565(2x)^1^1 --> from binomial theorem


    since 13 T3 +10T4 + T5 =0


    so  12285 (2x)^1^1 (4x^2 -20x +9) =0


    so x= zero or 0.5 or 4.5 since 2x^1^1 =0 or  4x^2-20x+9=0
    Last edited by mido22; October 13th 2012 at 03:26 AM.
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  4. #4
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    Re: two problems in binomial theorem

    2) t14 : t15 = 1:4 or 4:1 because he didn't say in the problem the between t14 to t15 or vice versa


    27C13 . x^1^4 : 27C14 . x^1^3 =1:4 or 4:1


    so  27C13 . x^1^4 : 27C13 . x^1^3 =1:4 or 4:1


    so x:1 = 1:4 or 4:1


    so x=4 or 0.25
    Last edited by mido22; October 13th 2012 at 03:26 AM.
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  5. #5
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    Re: two problems in binomial theorem

    For 1) I don't know why they didn't have 0 as a solution: you're answer looks spot on.

    For 2) I think they mean successive terms so by this, they mean that the next term is 4 times greater than the previous one.

    So basically divide T14/T13 = 4/1 = 4 which gives x = 4.

    This is just my interpretation though.

    You should ask your teacher for clarification, especially for number 1 in which you gave a good answer. Even your answer for question 2 is a solid one as well.
    Thanks from mido22
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  6. #6
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    Re: two problems in binomial theorem

    Hello, mido22!

    2) If the ratio between the two middle terms in the expansion of (1+x)^{27} is 1:4
    find the value of x.

    The two middle terms are the 14th and 15th: . {27\choose13}x^{13}\text{ and }{27\choose14}x^{14}

    The ratio is: . \frac{{27\choose13}x^{13}}{{27\choose14}x^{14}} \:=\:\frac{1}{4} \quad\Rightarrow\quad \frac{1}{x} \:=\:\frac{1}{4} \quad\Rightarrow\quad x \:=\:4


    I assumed that the ratio was in that order: 14th to the 15th.
    But I agree that the wording could have been more specific.

    I would have allowed full credit (maybe more) for your answer.
    It certainly demonstrates innovative thinking.
    Thanks from mido22
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  7. #7
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    Re: two problems in binomial theorem

    Thanks (Soroban and chiro) very much
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