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

1. anonymous

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

sorry, i'm still a bit confused.

3. mathmate

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

4. rishavraj

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

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

6. butterflydreamer

@mathmate the question is: $(3-x) (2x+1)^{10}$

7. mathmate

Thank you! You can continue with @rishavraj or @SithsAndGiggles

8. rishavraj

@butterflydreamer wht would be the 6th and 5th term??

9. butterflydreamer

the 5th term would be...... $\left(\begin{matrix}10 \\ 5\end{matrix}\right) (2x)^5 \times (3-x) ?$

10. rishavraj

yup :))) u see so in this case hmmm the coefficient of x would be -8064 ????

11. anonymous

\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

Ohhhhhh right right! I get it now, thank you :))! I got confused with the expansion :S