Coefficient of $x^4$ in multinomial expansion

Question

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)?

Answer 1

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.

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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.

Answer 2

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

Answer 3

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.

Answer 4

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: $$\begin{align}(1+x-2x^{2})^{7} &= \left(1 + \binom{7}{1}(2x) + \binom{7}{2}(2x)^2 + \binom{7}{3}(2x)^3 + \binom{7}{4}(2x)^4\right) \\ &\times\left(1 + \binom{7}{1}(-x) + \binom{7}{2}(-x)^2 + \binom{7}{3}(-x)^3 + \binom{7}{4}(-x)^4\right)\\ &= \left(1 + 14x + 84x^2 + 280x^3 + 560x^4 + ...\right)\\ &\times\left(1 - 7x + 21x^2 - 35x^3 + 35x^4 + ...\right)\\ &= (1\times35 + 14\times-35 + 84\times21 + 280\times-7 + 560\times1)x^4 + ...\\ &= -91x^4 + ... \end{align}$$