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What is the coefficient of $x^4$ in $(1 + x - 2x^2)^7$? What is a quick way to solve this problem using the binomial theorem (I have not learned multinomial theorem)?

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  • $\begingroup$ Why do you have to use the binomial theorem here? Using distributive law (and some combinatorics, incl. some binomial coefficients) would require less calculation. $\endgroup$ – arne.b Jan 24 '14 at 15:42
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An alternative approach, if you absolutely, positively have to use the binomial theorem, would be to let $a=1+x$ and $b=-2x^2$. Note that in the expansion of $(a+b)^7$ only the terms $a^7$, $a^6b$ and $a^5b^2$ will, when expanded again, eventually contain $x^4$. But how many? $a^7 = (x+1)^7$ should be easy, $a^6b=-2x^2(x+1)^6$,so you will need the coefficient of $x^2$ in $(x+1)^6$, and $a^5b^2=4x^4(1+x)^5$ should be easier again.


If you can do without the binomial theorem, fewer calculations will be required, however.

$(1 + x - 2x^2)^7=(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)(1 + x - 2x^2)$

The full product is the sum of all products of each combination of one summand from each expression in parentheses (not sure I put that in the best possible way; I mean: simply apply distributive law). Which of them contain $x^4$, and how many are there?

  • $1*1*1*x*x*x*x$ -- $\binom{7}{4}$ possibilities
  • $1*1*1*1*(-2x^2)*x*x$ -- $7*\binom{6}{2}$ possibilities (or $\binom{7}{2}*5$)
  • $1*1*1*1*1*1*(-2x^2)*(-2x^2)$ -- $\binom{7}{2}$ possibilities

Now calculate $\binom{7}{4}-2*7*\binom{6}{2}+(-2)(-2)\binom{7}{2}$ (note: This is also the expression you get with the former approach, and the result will be the one given in another answer already ;-)). I am not sure factorization saves much time here.

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  • $\begingroup$ Superb answer, sir!! $\endgroup$ – user34304 Jan 25 '14 at 3:30
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By factorizing the inner expresion, $$(1+x-2x^{2})^{7}=((1+2x)(1-x))^{7}=(1+2x)^{7}(1-x)^{7}$$ Using the Binomial Expansion on both, we get: $$\begin{align}(1+x-2x^{2})^{7} &= \left(1 + \binom{7}{1}(2x) + \binom{7}{2}(2x)^2 + \binom{7}{3}(2x)^3 + \binom{7}{4}(2x)^4\right) \\ &\times\left(1 + \binom{7}{1}(-x) + \binom{7}{2}(-x)^2 + \binom{7}{3}(-x)^3 + \binom{7}{4}(-x)^4\right)\\ &= \left(1 + 14x + 84x^2 + 280x^3 + 560x^4 + ...\right)\\ &\times\left(1 - 7x + 21x^2 - 35x^3 + 35x^4 + ...\right)\\ &= (1\times35 + 14\times-35 + 84\times21 + 280\times-7 + 560\times1)x^4 + ...\\ &= -91x^4 + ... \end{align}$$

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Hint: $$(1+x-2x^{2})^{7}=((1+2x)(1-x))^{7}=(1+2x)^{7}(1-x)^{7}$$

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  • $\begingroup$ Yes, but then what? $\endgroup$ – user34304 Jan 24 '14 at 14:44
  • $\begingroup$ You'll then have the product of two polynomials, $(a_{0}+a_{1}x+a_{2}x^{2}+\ldots)(b_{0}+b_{1}x+b_{2}x^{2}+\ldots)$. Which products of terms will give you an $x^{4}$? $\endgroup$ – preferred_anon Jan 24 '14 at 15:28
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Well, if you want to avoid multinomial theorem, you can start by noting that $$1+x-2x^2=1+2x-x-2x^2=1+2x-x(1+2x)=(1-x)(1+2x).$$ Hence, we can use binomial expansion twice on $$(1+x-2x^2)^7=(1-x)^7(1+2x)^7,$$ then find the $4$th degree term of the product of the two expanded binomials.

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  • $\begingroup$ How do u do that, sir? $\endgroup$ – user34304 Jan 24 '14 at 14:45
  • $\begingroup$ Which part do you mean? $\endgroup$ – Cameron Buie Jan 24 '14 at 14:48
  • $\begingroup$ The last part, sir $\endgroup$ – user34304 Jan 24 '14 at 14:58
  • $\begingroup$ Ah. Well, suppose that $$(1-x)^7=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots+a_7x^7$$ and $$(1+2x)^7=b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+\cdots+b_7x^7.$$ To get a degree $4$ term from a product of two monomial terms, they must be of degrees that add up to $4.$ There are only $5$ such pairs of terms in the product, and when we add them up, we get the degree $4$ term. That is, $$(1-x)^7(1+2x)^7=\text{LDS}+a_0b_4x^4+a_1b_3x^4+a_2b_2x^4 +a_3b_1x^4+a_4b_0x^4+\text{HDS}$$ (where LDS stand for "lower degree stuff," HDS stands for "higher degree stuff"), so the term is $$(a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0)x^4.$$ $\endgroup$ – Cameron Buie Jan 24 '14 at 15:11

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