Find the coefficient of \(x^{6}\) in
\[(1-2x+3x^{2}-4x^{3}+5x^{4})^{3}.\]
You should set out your working clearly.
By considering the binomial expansions of \((1+x)^{-2}\) and \((1+x)^{-6}\), or otherwise, find the
coefficient of \(x^{6}\) in
\[(1-2x+3x^{2}-4x^{3}+5x^{4}-6x^{5}+7x^{6})^{3}.\]
Solution:
We can obtain a \(6\) from \(4+2+0, 4+1+1, 3+3+0, 3+2+1, 2+2+2\).
So \(x^6\) from \(4,2,0\) can happen in \(6\) ways and gets us a coefficient of \(1 \cdot 3 \cdot 5\).
\(x^6\) from \(4,1,1\) can happen in \(3\) ways and gets us a coefficient of \(5 \cdot (-2) \cdot (-2)\).
\(x^6\) from \(3,3,0\) can happen in \(3\) ways and gets us a coefficient of \((-4) \cdot (-4) \cdot 1\).
\(x^6\) from \(3,2,1\) can happen in \(6\) ways and gets us a coefficient of \((-4) \cdot 3 \cdot (-2)\).
\(x^6\) from \(2,2,2\) can happen in \(1\) ways and gets us a coefficient of \(3 \cdot 3 \cdot 3\).
This leaves us with a total coefficient of:
\(6 \cdot 15 + 3 \cdot 20 + 3 \cdot 16 + 6 \cdot 24 + 1 \cdot 27 = 369\)
\begin{align*}
(1+x)^{-2} &= 1 + (-2)x+\frac{(-2)\cdot(-3)}{2!} x^2 + \frac{(-2)(-3)(-4)}{3!}x^3 + \cdots \\
&= 1 -2x+3x^2-4x^3+5x^4+\cdots \\
\end{align*}
The coefficient of \(x^6\) in the expansion of \((1+x)^{-6}\) will be \(\frac{(-6)(-7)(-8)(-9)(-10)(-11)}{6!} = \frac{11!}{6!5!} = 462\).
The coefficient of \(x^6\) in the expansion of \((1 -2x+3x^2-4x^3+5x^4+\cdots)^3\) will be the same as the coefficient of \(x^6\) in the expansion of \((1 -2x+3x^2-4x^3+5x^4-6x^5+7x^6)^3\), ie it will be \(462\)