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.



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 $$\binom{n}{k}=\binom{n}{n-k}\ .$$

  • $\begingroup$ 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

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 $\binom{15}{8}(3p^2)^{8}(-2q)^{15-8}$.

  • $\begingroup$ 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

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