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Find the coefficient of $x^{17}$ in the expansion of $(3x^7 + 2x^5 -1)^{20}$

I'm stuck in handling this question as I do know how to solve it when it has 2 terms.
But now it has 3.

I have no idea where to begin...

Christopher Marley
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coffee
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    You can do it a number of ways, but if you don't want to have to learn something new, then recognize that $(3x^7+2x^5-1)^{20}=((3x^7+2x^5)-1)^{20}$ – JMoravitz May 04 '18 at 05:15
  • In how many ways can you write $17$ as a sum of $5$'s and $7$'s? – Jyrki Lahtonen May 04 '18 at 05:21
  • I tried using your method and got stuck at the second part where I had $(3x^7 + 2x^5)^{20-n}$

    I further expand the derived expression and had 2 unknowns

     – coffee May 04 '18 at 05:34
  • Let $y=(3x^7+2x^5)$. Making this substitution, we have the expression expands as $\binom{20}{0}(-1)^{20}y^0+\binom{20}{1}(-1)^{19}y^1+\binom{20}{2}(-1)^{18}y^2+\binom{20}{3}(-1)^{17}y^3+\dots$. Can you see how to continue? Can you recognize that $(3x^7+2x^5)^4,(3x^7+2x^5)^5,(3x^7+2x^5)^6,\dots$ each cannot possibly contribute to the final coefficient of $x^{17}$? Where then should all of your attention be drawn? – JMoravitz May 04 '18 at 05:45
  • @JMoravitz I understand thanks!. One query is that given the expansion requirements is small, I can manually do it out all right? – coffee May 04 '18 at 05:56

3 Answers3

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=the coefficient of $x^{17}$ in $(2x^5-1)^{20}$

$+\binom{20}13^1\cdot$ the coefficient of $x^{17-7}$ in $(2x^5-1)^{20-1}$

$+\binom{20}23^2\cdot$ the coefficient of $x^{17-2\cdot7}$ in $(2x^5-1)^{20-2}$

lab bhattacharjee
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  • How did you get to this process? Is it a fancy binomial expansion, or some really abstract math trick? – Christopher Marley May 04 '18 at 05:17
  • @ChristopherMarley it is essentially the same hint as I gave above, except having broken it up as $(3x^7+(2x^5-1))^{20}$ instead. Its just repeated application of binomial expansion. – JMoravitz May 04 '18 at 05:18
  • @JMoravitz I tried using your method and got stuck at the second part where I had (3x7+2x5)20−n I further expand the derived expression and had 2 unknowns. – coffee May 04 '18 at 05:41
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We can apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of an expression.

We obtain \begin{align*} \color{blue}{[x^{17}]}&\color{blue}{(3x^7+2x^5-1)^{20}}\\ &=[x^{17}]\sum_{k=0}^{20}\binom{20}{k}(3x^7)^k(2x^5-1)^{20-k}\tag{1}\\ &=\binom{20}{0}3^0[x^{17}](2x^5-1)^{20}+\binom{20}{1}3^1[x^{10}](2x^5-1)^{19}\\ &\qquad +\binom{20}{2}3^2[x^3](2x^5-1)^{18}\tag{2}\\ &=3\binom{20}{1}[x^{10}]\sum_{j=0}^{19}\binom{19}{j}(2x^5)^j(-1)^{19-j}\tag{3}\\ &=-3\binom{20}{1}\binom{19}{2}2^2\tag{4}\\ &\,\,\color{blue}{=-41\,040} \end{align*}

Comment:

  • In (1) we apply the binomial theorem once to $(3x^7+(2x^5-1))^{20}$.

  • In (2) we observe that the term $(3x^7)^k$ contributes to $[x^{17}]$ only when $k=0,1,2$. We also apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

  • In (3) we note that expanding $(2x^5-1)^n$ gives only exponents of $x$ which are a multiple of $5$. We can therefore stick to the middle summand with $[x^{10}]$ and we apply the binomial theorem again.

  • In (4) we select the coefficient of $x^j$ accordingly.

Markus Scheuer
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The trinomial expansion: $$(a+b+c)^n=\sum_{\ \ \ \ \ i,j,k\\ i+j+k=n}{n\choose {i,j,k}}a^ib^jc^k,$$ where ${n\choose {i,j,k}}=\frac{n!}{i!j!k!}$.

If applied to the given trinomial: $$(3x^7 + 2x^5 -1)^{20}=\sum_{\ \ \ \ \ i,j,k\\ i+j+k=20}{20\choose {i,j,k}}(3x^7)^i(2x^5)^j(-1)^k.$$ The exponent of $x$ will be $17$ when: $$7i+5j=17 \Rightarrow (i,j,k)=(1,2,17).$$ Hence: $${20\choose {1,2,17}}3^12^2(-1)^{17}=\frac{20!}{1!2!17!}(-12)=-41040.$$

farruhota
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