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butterflydreamer

  • one year ago

Find the coefficient of x^6 in the expansion of (3-x)(2x+1)^10

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  1. anonymous
    • one year ago
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    First consider the expansion of \((2x+1)^{10}\). By the binomial theorem, \[(2x+1)^{10}=\sum_{k=0}^{10}\binom {10}k(2x)^k1^{n-k}=\sum_{k=0}^{10}\binom {10}k(2x)^k\] The term containing \(x^5\) and \(x^6\) occur when \(k=5\) and \(k=6\), respectively, so let's grab those terms: \[\binom {10}5(2x)^5=\cdots\\ \binom{10}6(2x)^6=\cdots\] We pick these powers specifically because multiplying the \(x^5\) term by the \(-x\) in \(3-x\) generates a term with \(x^6\), while the \(x^6\) term gets multiplied by \(3\) in \(3-x\). The coefficient of the \(x^6\) term in \((3-x)(2x+1)^{10}\) is then the sum of the coefficients above.

  2. butterflydreamer
    • one year ago
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    sorry, i'm still a bit confused.

  3. mathmate
    • one year ago
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    @butterflydreamer Just to be clear about the question, is it [(3-x)(2x+1)]^10 or (3-x)[(2x+1)^10]

  4. rishavraj
    • one year ago
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    see u will get terms with x^6 in the 5th term and also the 6th term bcoz u gotta multiply it with (3 - x)

  5. rishavraj
    • one year ago
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    the 6th would have x^5 . bt all bcoz u gotta multiply with (3 - x) it would change to x^6

  6. butterflydreamer
    • one year ago
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    @mathmate the question is: \[(3-x) (2x+1)^{10}\]

  7. mathmate
    • one year ago
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    Thank you! You can continue with @rishavraj or @SithsAndGiggles

  8. rishavraj
    • one year ago
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    @butterflydreamer wht would be the 6th and 5th term??

  9. butterflydreamer
    • one year ago
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    the 5th term would be...... \[\left(\begin{matrix}10 \\ 5\end{matrix}\right) (2x)^5 \times (3-x) ?\]

  10. rishavraj
    • one year ago
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    yup :))) u see so in this case hmmm the coefficient of x would be -8064 ????

  11. anonymous
    • one year ago
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    \[\begin{align*}(2x+1)^{10}&=\sum_{k=0}^{10}\binom {10}k(2x)^k\\ &=\cdots+\binom{10}5(2x)^5+\binom{10}6(2x)^6+\cdots\\ &=\cdots+32\binom{10}5x^5+64\binom{10}6x^6+\cdots\end{align*}\] Multiplying by \(3-x\), you apply the distributive property: \[(3-x)\left(32\binom{10}5x^5+64\binom{10}6x^6\right)\\ \quad\quad=3\times64\binom{10}6x^6+(-x)\times 32\binom{10}5x^5+\text{other stuff}\\ \quad\quad =3\times64\binom{10}6x^6- 32\binom{10}5x^6+\text{other stuff}\\ \quad\quad =\left(3\times64\binom{10}6- 32\binom{10}5\right)x^6+\text{other stuff}\]

  12. butterflydreamer
    • one year ago
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    Ohhhhhh right right! I get it now, thank you :))! I got confused with the expansion :S

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