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Math Help - Binomial expansion

  1. #1
    Member ~berserk's Avatar
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    Binomial expansion

    what is the third term in the expansion of (a-3b)^4
    Also the middle term of the expansion (sinx +2)^4
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by ~berserk View Post
    what is the third term in the expansion of (a-3b)^4
    Also the middle term of the expansion (sinx +2)^4

    The binomial theorem states that

    (a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k

    Hint the first term is when n=0.

    So what would the third term be.

    I hope this helps.
    Good luck.
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  3. #3
    Member ~berserk's Avatar
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    I just have one more question. For example 5th term in (3a-b)^6 i get (15)(3)^2a^2b^4 but it is -b does that make a difference or is the answer 135a^2b^4
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by ~berserk View Post
    I just have one more question. For example 5th term in (3a-b)^6 i get (15)(3)^2a^2b^4 but it is -b does that make a difference or is the answer 135a^2b^4
    In general yes, in this case no. Here is why.

    Consider

    (a-b)^5=[a+(-b)]^5
    =a^5(-b)^0+5a^4(-b)^1+10a^3(-b)^2+10a^2(-b)^3+5(a)^1(-b)^4+a^0(-b)^5=

    a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5

    I hope this helps
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  5. #5
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    Expand x.sin2x upto the term containing x7

    Any Ideas, i'm sooooooo stuck
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Squeeka7 View Post
    Any Ideas, i'm sooooooo stuck
    You wrote this

    x\sin(2x)=2x\sin(x)\cos(x)

    Did you mean (x+\sin(2x))^7
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  7. #7
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    Erm no, to give a bit of back ground the question is part of an Analytical methods assignment, and is written "Expand x.sin2x upto the term containing x7"
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Squeeka7 View Post
    Erm no, to give a bit of back ground the question is part of an Analytical methods assignment, and is written "Expand x.sin2x upto the term containing x7"
    Well then I assume it is a Maclaurin series you want

    x\sin(2x)=\sum_{n=0}^{\infty}\frac{(-1)^n(2x)^{2n+2}}{(2n+1)!}

    So what you want is this

    \sum_{n=0}^{3}\frac{(-1)^n(2x)^{2n+2}}{(2n+1)!}

    But there will be no 7 term, a 6 and 8 term though

    P.S. This is hardly an Algebra/Pre-algebra question.
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