2
$\begingroup$

I am just trying to understand why the term is $\binom{15}8$(3p$^2$ - 2q)$^7$.

I need to find the coefficient in $p^{16}q^7$ in $(3p^2 - 2q)^{15}$

So, I know that $n = 15$ and I have $a^{n - k}b^k$ but I cannot figure out how to get $\binom{15}8$. I've tried watching videos on YouTube and looking up some tutorials but I only found them confusing. I don't have much useful work to add here. The book gives an answer but no explanation and I am completely stuck trying to figure out this.

I am not looking for the complete answer. Just enough to get $\binom{15}8$.

Thanks for any help.

Tony

$\endgroup$
3
$\begingroup$

You will need the binomial expansion $$(a+b)^{15}=a^{15}+\binom{15}1a^{14}b+\cdots+\binom{15}{k}a^{15-k}b^k+\cdots+b^{15}\ .$$ Substituting $a=3p^2$ and $b=-2q$ gives $$(3p^2-2q)^{15}=\cdots+\binom{15}{k}(3p^2)^{15-k}(-2q)^k+\cdots\ .$$ Now look at the general term $$\binom{15}{k}(3p^2)^{15-k}(-2q)^k$$ and work out what value of $k$ you need in order to get a $p^{16}q^7$ term; the coefficient of this term is $$\binom{15}{k}3^{15-k}(-2)^k$$ and this will be your answer.

To match the given answer you will also need to use the fact that $$\binom{n}{k}=\binom{n}{n-k}\ .$$

$\endgroup$
  • $\begingroup$ Hi, thanks for the feed back but I'm afraid I still don't know how to figure this out. Specifically, I cannot figure this part out: d work out what value of k you need in order to get a $p^16q^7$ term; the coefficient of this term is $\binom{15}{k}3^{15−k}(−2)^k$ $\endgroup$ – Tony Mar 20 '14 at 1:43
  • $\begingroup$ If you expand $\binom{15}{k}(3p^2)^{15-k}(-2q)^k$, you will get a coefficient times a power of $p$ times a power of $q$. What (in terms of $k$) is the power of $p$? What is the power of $q$? What is the coefficient? Can you find a value of $k$ which makes the powers match what you want, namely $p^{16}$ and $q^7$? If so, then substituting this value of $k$ into the coefficient gives the answer. $\endgroup$ – David Mar 20 '14 at 1:50
  • $\begingroup$ Hi, I see where I was going wrong now. Once I got this $\binom{15}{8}(3p^2)(-2q)^{15-8}$ I was trying to solve this equation and did not realize that when I had $p^{16}$ and $q^7$ I had my answer and that this is why k is equal to 8. $\endgroup$ – Tony Mar 20 '14 at 1:53
1
$\begingroup$

We have the following formula: $$(x+y)^{n}=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k} $$ called the binomial formula. When $k=8$ we have (take $x=3p^2$ and $y=-2q$ in the binomial formula) the term $\binom{15}{8}(3p^2)^{8}(-2q)^{15-8}$.

$\endgroup$
  • $\begingroup$ Hi, thanks for the feedback. I really appreciate it. Between your response and David's I finally figured out where I was going wrong. I'll do more of these exercises to reenforce this topic. $\endgroup$ – Tony Mar 20 '14 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.