I know that I should first factor the expression within the brackets to $-(3x+1)^6(x-1)^6$. However, after that I do not know what to do.
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1We can use General Leibniz rule for taking $5$th derivative and it's value at $x=0$ will be $5!$ times the desired value. – Alexey Burdin Jul 08 '20 at 02:35
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5Use binomial theorem to expand those two sixth powers, then pick out the terms that give $x^5$ when you multiply out. – Gerry Myerson Jul 08 '20 at 02:35
3 Answers
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Actually you can do it without factorization as well. Just like binomial theorem, we have a multinomial theorem as well. For example, $$(a+b+c)^n=\sum_{\substack{i,j,k\\i+j+k=n}}\frac{n!}{i!j!k!}a^ib^jc^k.$$ This can be generalized to higher order multinomials as well.
So in your case, $$(1+2x-3x^2)^6=\sum_{\substack{i,j,k\\i+j+k=6}}\frac{6!}{i!j!k!}(1)^i\,(2x)^j\,(-3x^2)^k=\sum_{\substack{i,j,k\\i+j+k=6}}\frac{6!}{i!j!k!}2^j(-3)^k\,x^{\color{red}{j+2k}}.$$ For the coefficient of $x^5$, we want $j+2k=5$ and $i+j+k=6$. This means $i+(5-k)=6 \implies i-k=1.$ Thus $$(i,j,k) \in \{(1,5,0), (2,3,1), (3,1,2)\}.$$ Thus the coefficient of $x^5$ is $$\frac{6!}{1!5!0!}2^5(-3)^0+\frac{6!}{2!3!1!}2^3(-3)^1+\frac{6!}{3!1!2!}2^1(-3)^2=\color{blue}{-168}.$$
Anurag A
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1Try to give hints first rather than posting complete solution. Read this. – SarGe Jul 08 '20 at 05:07
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@SarGe Mothing against the issue you have raised but hints were already given for the question based on OP's approach. This is a different method and that is why I had added more details. – Anurag A Jul 08 '20 at 09:34
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@AnuragA, I agree. But still you can also give the key statements of your method and let OP think over it. – SarGe Jul 08 '20 at 09:39
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@SarGe I made a judgement call for that plus there was a conversation between another user JCAA and me regarding certain aspects of the answer I gave (which both of us have mostly deleted, barring two that are left (see comments)) that necessitated the addition of details. – Anurag A Jul 08 '20 at 09:43
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@SarGe You can see reminiscences of it above. But then if you cannot see something then it doesn't mean it never existed. In which case rushing to judgement is also not good. – Anurag A Jul 08 '20 at 09:51
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I agree my mistake but what I personally feel that this site is not for one-to-one correspondence and so the conversation should not have been deleted as it might be helpful for future audience. – SarGe Jul 08 '20 at 09:58
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There shouldn't be a minus sign outside, it's wiped out when you take the sixth power.
$(1+2x-3x^2)^6 = (3x+1)^6(x-1)^6$
Now just find all the ways you can get to a power $5$ in the result. $5 = 0+5 = 1+4 = 2+3 = 3+2 = 4+1 = 5+0$.
So you need to find only the relevant terms in each of the individual expansions. As an example here is "$3+2$":
$\binom 6 3 (3x)^3(1)^3 \cdot \binom 6 2(x)^2(-1)^4$.
Compute all similar terms and add them up.
Deepak
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We can also apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ of an expression.
We obtain \begin{align*} \color{blue}{[x^5]}&\color{blue}{(1+2x-3x^2)^{6}}\\ &=[x^5]\sum_{j=0}^{6}\binom{6}{j}(-3x^2)^j(1+2x)^{6-j}\tag{1}\\ &=\sum_{j=0}^2\binom{6}{j}(-3)^j[x^{5-2j}]\sum_{k=0}^{6-j}\binom{6-j}{k}(2x)^k\tag{2}\\ &=\sum_{j=0}^2\binom{6}{j}(-3)^j\binom{6-j}{5-2j}2^{5-2j}\tag{3}\\ &=\binom{6}{0}(-3)^0\binom{6}{5}2^5+\binom{6}{1}(-3)^1\binom{5}{3}2^3+\binom{6}{2}(-3)^2\binom{4}{1}2^1\\ &=192-1\,440+1\,080\\ &\,\,\color{blue}{=-168} \end{align*}
Comment:
In (1) we apply the binomial theorem once to $((1+2x)-3x^2))^6$.
In (2) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$ and we set the upper index of the outer sum to $2$, since other terms do not contribute. We also apply the binomial theorem once again.
In (3) we select the coefficient of $x^{5-2j}$.
Markus Scheuer
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