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Thread: Coefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is?

  1. #1
    Super Member fardeen_gen's Avatar
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    Coefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is?

    Coefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is -

    A) 5C2
    B) ∑(r = 0 to 4) (-1)^r 6Cr . 5C(4 - r)
    C) - 5C3
    D) 5C4

    More than one options may be correct.
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  2. #2
    MHF Contributor red_dog's Avatar
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    The general term of $\displaystyle (1+x)^5$ is $\displaystyle C_5^kx^k, \ 0\leq k\leq 5$

    The general term of $\displaystyle (1-x)^6$ is $\displaystyle (-1)^lC_6^lx^l, \ 0\leq l\leq 6$

    Then the genaral term of the entire expansion is $\displaystyle (-1)^lC_5^kC_6^lx^{k+l}, \ 0\leq k\leq 5, \ 0\leq l\leq 6$

    So the condition is $\displaystyle k+l=4$.

    Now take all the possibilities for k and l.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Coefficient of x^4 in the product of the expansions (1 + x)^5.(1 - x)^6 is -

    A) 5C2
    B) ∑(r = 0 to 4) (-1)^r 6Cr . 5C(4 - r)
    C) - 5C3
    D) 5C4

    More than one options may be correct.
    Hint:

    $\displaystyle {\left( {1 + x} \right)^5}{\left( {1 - x} \right)^6} = {\left( {1 + x} \right)^5}{\left( {1 - x} \right)^5}\left( {1 - x} \right) = {\left( {1 - {x^2}} \right)^5}\left( {1 - x} \right) = {\left( {{x^2} - 1} \right)^5}\left( {x - 1} \right)$
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  4. #4
    Super Member fardeen_gen's Avatar
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    The general term of is

    The general term of is

    Then the genaral term of the entire expansion is

    So the condition is .

    Now take all the possibilities for k and l.
    Correct me if I am wrong. 5C4 is the only possible option.
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  5. #5
    MHF Contributor red_dog's Avatar
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    No.
    You have the possibilities:
    k=0, l=4
    k=1, l=3
    k=2, l=2
    k=3, l=1
    k=4, l=0

    Then the coefficient is

    $\displaystyle C_5^0C_6^4-C_5^1C_6^3+C_5^2C_6^2-C_5^3C_6^1+C_5^4C_6^0=\sum_{r=0}^4(-1)^rC_6^rC_5^{4-r}$
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  6. #6
    Senior Member DeMath's Avatar
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    Members of the decomposition of the factor $\displaystyle {\left( {{x^2} - 1} \right)^5}$ have only even powers. Therefore, you just enough to consider this factor.

    Your answer must be $\displaystyle C_5^3 = 10$ for $\displaystyle x^4$.
    Last edited by DeMath; Feb 14th 2009 at 12:08 AM.
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  7. #7
    Super Member fardeen_gen's Avatar
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    So correct answers are:

    B), C) and D), right?
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  8. #8
    MHF Contributor red_dog's Avatar
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    No, the correct answer is B.
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  9. #9
    Senior Member DeMath's Avatar
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    $\displaystyle {\left( {{x^2} - 1} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{x^{2\left( {5 - k} \right)}}{{\left( { - 1} \right)}^k}} = C_5^0{x^{10}} - C_5^1{x^8} + C_5^2{x^6} - C_5^3{x^4} + C_5^4{x^2} - C_5^5{x^0}$

    $\displaystyle {\left( {1 + x} \right)^5}{\left( {1 - x} \right)^6} = {\left( {{x^2} - 1} \right)^5}\left( {x - 1} \right) \Rightarrow C_5^3 = 10{\text{ for }}{x^4}$

    Sorry, I corrected.
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  10. #10
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by DeMath View Post
    Only C)

    $\displaystyle {\left( {{x^2} - 1} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{x^{2\left( {5 - k} \right)}}{{\left( { - 1} \right)}^k}} = C_5^0{x^{10}} - C_5^1{x^8} + C_5^2{x^6} - C_5^3{x^4} + C_5^4{x^2} - C_5^5{x^0}$

    $\displaystyle (1+x)^5(1-x)^6\neq (x^2-1)^5$ !!!
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  11. #11
    Super Member fardeen_gen's Avatar
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    red_dog, A) is equivalent to B). So A) is also correct, right?

    And aren't we supposed to take the individual possibility of k = 4 and l = 0, which gives 5C4(option D)?
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