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Math Help - Binomial expansion

  1. #1
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    Binomial expansion

    Find the coefficients of x^7 in the expansion of (1-3x^2)(x^2-\frac{2}{x})^7

    My attempt:

    to obtain x^7, coefficients of x^5 and x^7 have to be used.

     T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r

    therefore, 2(7-r)+(-r)=5
    r=3


    OR 2(7-r)+(-r)=5
    r= 2(\frac{1}{3})

    Puzzled at arriving at 2\frac{1}{3} for r.
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  2. #2
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    Quote Originally Posted by Punch View Post
    Find the coefficients of x^7 in the expansion of (1-3x^2)(x^2-\frac{2}{x})^7

    My attempt:

    to obtain x^7, coefficients of x^5 and x^7 have to be used.

     T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r

    therefore, 2(7-r)+(-r)=5
    r=3


    OR 2(7-r)+(-r)=5
    r= 2(\frac{1}{3})

    Puzzled at arriving at 2\frac{1}{3} for r.
    is the answer 840 ?
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    is the answer 840 ?
    yes it is! please show me the right way of solving it
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  4. #4
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    Quote Originally Posted by Punch View Post
    yes it is! please show me the right way of solving it
    Great . Of course , we can expand the whole thing and combine terms which will produce x^7 but it would be tedious and risky so its better to use the binomial theorem :

     <br />
(a+b)^n=\sum^{n}_{k=0} \binom{n}{k} a^{n-k}b^k<br />

     <br />
(x^2-\frac{2}{x})^7=\sum^{n}_{k=0} \binom{7}{k}x^{14-3k}\cdot 2^k<br />

    (1-3x^2)(x^2-\frac{2}{x})^7

    =\sum^{n}_{k=0} \binom{7}{k}x^{14-3k}\cdot 2^k+\sum^{n}_{k=0}\binom{7}{k} x^{16-3k}\cdot 3\cdot 2^k

    so now lets see what values of k give 7

    14-3k=7 , k=7/3 and this cant be since k must be +ve integer

    how about 16-3k=7 , k=3 .

    its coefficient is \binom{7}{3} \cdot 3 \cdot 2^3=840
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  5. #5
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    I didnt quite understand "so now lets see what values of k give 7

    14-3k=7 , k=7/3 and this cant be since k must be +ve integer

    how about 16-3k=7 , k=3 ."

    Does this shows that if the value of r is not a whole number, we have to round the number off and see if the power of x is what we are after?
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  6. #6
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    Quote Originally Posted by Punch View Post
    Find the coefficients of x^7 in the expansion of (1-3x^2)(x^2-\frac{2}{x})^7

    My attempt:

    to obtain x^7, coefficients of x^5 and x^7 have to be used.

    \color{blue}x^0 and \color{blue}x^7, \color{blue}x^2 and \color{blue}x^5

     T_{r+1}= (7Cr)(x^2)^{7-r}(\frac{-2}{x})^r

    the above is fine if you recognise that there is no \color{blue}x^7 term when you multiply the binomial expansion by 1

    \color{blue}x^{14-2r}x^{-r}=x^{14-3r}=x^7 has no natural solution


    Then the \color{blue}x^7 term is obtained by multiplying the binomial expansion by \color{blue}-3x^2


    therefore, 2(7-r)+(-r)=5
    r=3


    OR 2(7-r)+(-r)=5
    r= 2(\frac{1}{3}) you should have 14-2r-r=14-3r=5, 3r=14-5=9

    Puzzled at arriving at 2\frac{1}{3} for r.
    check your 2nd last line simplification again

    .
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  7. #7
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    alright so can i conclude that I only have to use the term coefficient of x^5 since there is no x^7 term
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  8. #8
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    Quote Originally Posted by Punch View Post
    alright so can i conclude that I only have to use the term coefficient of x^5 since there is no x^7 term
    Yes, it boils down to finding the x^5 term of the binomial.

    Then the coefficient of x^7 is -3(coefficient\ of\ x^5\ from\ the\ binomial\ expansion)

    This is a shortcut to multiplying out all the terms independently,
    if you only want a single term of the entire expansion of the factors.
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