Determine the first three terms in the expansion of \({(1 - 2x)^5}{(1 + x)^7}\) in ascending powers of x.

METHOD 1

constant term: \(\left( {\begin{array}{*{20}{c}}
  5 \\
  0
\end{array}} \right){( - 2x)^0}\left( {\begin{array}{*{20}{c}}
  7 \\
  0
\end{array}} \right){x^0} = 1\)     A1

term in x: \(\left( {\begin{array}{*{20}{c}}
  7 \\
  1
\end{array}} \right)x + \left( {\begin{array}{*{20}{c}}
  5 \\
  1
\end{array}} \right)( - 2x) = - 3x\)     (M1)A1

term in \({x^2}\) : \(\left( {\begin{array}{*{20}{c}}
  7 \\
  2
\end{array}} \right){x^2} + \left( {\begin{array}{*{20}{c}}
  5 \\
  2
\end{array}} \right){( - 2x)^2} + \left( {\begin{array}{*{20}{c}}
  7 \\
  1
\end{array}} \right)x\left( {\begin{array}{*{20}{c}}
  5 \\
  1
\end{array}} \right)( - 2x) = - 9{x^2}\)     M1A1     N3

[5 marks]

METHOD 2

\({(1 - 2x)^5}{(1 + x)^7} = \left( {1 + 5( - 2x) + \frac{{5 \times 4{{( - 2x)}^2}}}{{2!}} + ...} \right)\left( {1 + 7x + \frac{{7 \times 6}}{2}{x^2} + ...} \right)\)     M1M1

\( = (1 - 10x + 40{x^2} + ...)(1 + 7x + 21{x^2} + …)\)

\( = 1 + 7x + 21{x^2} - 10x - 70{x^2} + 40{x^2} + …\)

\( = 1 - 3x - 9{x^2} + …\)     A1A1A1     N3

[5 marks]

Although the majority of the candidates understood the question and attempted it, excessive time was spent on actually expanding the expression without consideration of the binomial theorem. A fair amount of students confused “ascending order”, giving the last three instead of the first three terms.