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1025867@etusd.org

 
 
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1025867@etusd.org
by shawn luo - Saturday, February 8, 2020, 11:43 AM
 

how to solve for coefficents

 
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Re: 1025867@etusd.org
by Areteem Professor - Monday, February 10, 2020, 11:23 AM
 

Hi Shawn.

Could you be a little more specific? Is there a problem you are trying to solve?

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Re: 1025867@etusd.org
by shawn luo - Wednesday, February 12, 2020, 4:14 PM
 

I meant like x^7 in (x-3)^10 as in problem 8.22

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Re: 1025867@etusd.org
by Areteem Professor - Thursday, February 13, 2020, 10:40 AM
 

Just remember to use the Binomial Theorem.

The binomial Theorem tells us that $$\begin{aligned}(x-3)^{10} &=  \sum_{k = 0}^{10} \binom{10}{k} x^{n-k} \cdot (-3)^k \\ &= \binom{10}{0} x^{10} + \binom{10}{1} x^{9}\cdot (-3) + \cdots + \binom{10}{9} x\cdot (-3)^9 + \binom{10}{10} (-3)^{10}\end{aligned}$$ so the term with $x^7$ in the expansion of $(x - 3)^{10}$ is $\displaystyle \binom{10}{3} x^7 \cdot (-3)^3$, so the coefficient of $x^7$ is $\displaystyle \binom{10}{3} \cdot (-3)^3 = -3240$.