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butterflydreamer
 one year ago
Find the coefficient of x^6 in the expansion of (3x)(2x+1)^10
butterflydreamer
 one year ago
Find the coefficient of x^6 in the expansion of (3x)(2x+1)^10

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First consider the expansion of \((2x+1)^{10}\). By the binomial theorem, \[(2x+1)^{10}=\sum_{k=0}^{10}\binom {10}k(2x)^k1^{nk}=\sum_{k=0}^{10}\binom {10}k(2x)^k\] 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 \(3x\) generates a term with \(x^6\), while the \(x^6\) term gets multiplied by \(3\) in \(3x\). The coefficient of the \(x^6\) term in \((3x)(2x+1)^{10}\) is then the sum of the coefficients above.

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i'm still a bit confused.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0@butterflydreamer Just to be clear about the question, is it [(3x)(2x+1)]^10 or (3x)[(2x+1)^10]

rishavraj
 one year ago
Best ResponseYou've already chosen the best response.0see u will get terms with x^6 in the 5th term and also the 6th term bcoz u gotta multiply it with (3  x)

rishavraj
 one year ago
Best ResponseYou've already chosen the best response.0the 6th would have x^5 . bt all bcoz u gotta multiply with (3  x) it would change to x^6

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate the question is: \[(3x) (2x+1)^{10}\]

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Thank you! You can continue with @rishavraj or @SithsAndGiggles

rishavraj
 one year ago
Best ResponseYou've already chosen the best response.0@butterflydreamer wht would be the 6th and 5th term??

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0the 5th term would be...... \[\left(\begin{matrix}10 \\ 5\end{matrix}\right) (2x)^5 \times (3x) ?\]

rishavraj
 one year ago
Best ResponseYou've already chosen the best response.0yup :))) u see so in this case hmmm the coefficient of x would be 8064 ????

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

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0Ohhhhhh right right! I get it now, thank you :))! I got confused with the expansion :S
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