Online Course Discussion Forum

1025867@etusd.org

 
 
luoshawn的头像
1025867@etusd.org
luoshawn - 2020年02月8日 Saturday 11:43
 

how to solve for coefficents

 
ProfessorAreteem的头像
Re: 1025867@etusd.org
ProfessorAreteem - 2020年02月10日 Monday 11:23
 

Hi Shawn.

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

luoshawn的头像
Re: 1025867@etusd.org
luoshawn - 2020年02月12日 Wednesday 16:14
 

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

ProfessorAreteem的头像
Re: 1025867@etusd.org
ProfessorAreteem - 2020年02月13日 Thursday 10:40
 

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 $\displaystyle \binom{10}{3} \cdot (-3)^3 = -3240$.