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Math Help - Binomial theorem question check

  1. #1
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    Binomial theorem question check

    hey guys, just need to check if my working out and solution is correct.

    Find the coefficient of x^6 in the expansion (x + 3)^8

    My solution


    So: \sum_{i=0}^{8} \binom{8}{8-i} (x)^{8-i} (3)^i

    now since we need to find coefficient of x^6, therefore  i = 6.

    so \binom{8}{2} (1)^2 (3)^6 = 28 \times 1 \times 739 = 28 \times 739
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  2. #2
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    Quote Originally Posted by jvignacio View Post
    hey guys, just need to check if my working out and solution is correct.

    Find the coefficient of x^6 in the expansion (x + 3)^8

    My solution


    So: \sum_{i=0}^{8} \binom{8}{8-i} (x)^{8-i} (3)^i

    now since we need to find coefficient of x^6, therefore  i = 6.

    so \binom{8}{2} (1)^2 (3)^6 = 28 \times 1 \times 739 = 28 \times 739
    It's not.

    The general term is {8 \choose r} x^{8-r} 3^r. You require the coefficient corresponding to 8 - r = 6.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    It's not.

    The general term is {8 \choose r} x^{8-r} 3^r. You require the coefficient corresponding to 8 - r = 6.
    So: \sum_{i=0}^{8} \binom{8}{i} (x)^{8-i} (3)^i

    now since we need to find coefficient of x^6,  8 - i = 6 , so i = 2..

    so \binom{8}{2} (1)^6 (3)^2 = 28 \times 1 \times 9 = 28 \times 9 = 252

    this better?
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