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    hey

    I have came across an unfamiliar binomial question which I am unsure of

    "Find the term independent of x in the binomial expansion of

    (3x + 1/x2)9"

    How would I do this? Thanks
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    (Original post by upthegunners)
    hey

    I have came across an unfamiliar binomial question which I am unsure of

    "Find the term independent of x in the binomial expansion of

    (3x + 1/x2)9"

    How would I do this? Thanks
    The eaxpansion means szum of 10 terms (from k=0 ..9)
    that is
    \displaystyle \sum_{k=0}^9 \binom{9}{k} \cdot \left (3x\right )^{9-k}\cdot \left (\frac{1}{x^2}\right )^k

    for k th term
    \displaystyle \binom{9}{k} \cdot 3^{9-k}\cdot x^{9-k}\cdot \frac{1}{x^{2k}}
    This independent from x when 9-k=2k -> k=3
    The 4th term.
    the value of it
    \displaystyle \binom{9}{3}\cdot 3^6 =\frac{9\cdot 8 \cdot 7}{3\cdot 2}\cdot 3^6=3^7\cdot 28=61236
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    The first term will be

    (number) (3x)^9

    The next term will be

    (number) (3x)^8 (1/x^2)^1 = (number) (x^6)

    Continue this until you find the one with just a number

    Then work out the coefficient
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    (Original post by TenOfThem)
    The first term will be

    (number) (3x)^9

    The next term will be

    (number) (3x)^8 (1/x^2)^1 = (number) (x^6)

    Continue this until you find the one with just a number

    Then work out the coefficient
    Is there not a quicker method?
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    (Original post by upthegunners)
    Is there not a quicker method?
    Realising that there is an obvious pattern and that you need

    (number) (3x)^a (1/x^2)^b

    where a+b = 9

    and a-2b = 0

    ie post 2

    however, I went for a simpler approach based on my understanding of your capabilities
 
 
 
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