Write down the number of terms in this expansion.

Find the term in \({x^3}\).

9 terms     A1     N1

[1 mark]

valid approach to find the required term     (M1)

eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 8 \\ r \end{array}} \right){(2x)^{8 - r}}{(3)^r},{\text{ }}{(2x)^8}{(3)^0} + {(2x)^7}{(3)^1} +  \ldots \), Pascal’s triangle to \({{\text{8}}^{{\text{th}}}}\) row

identifying correct term (may be indicated in expansion)     (A1)

eg\(\;\;\;{{\text{6}}^{{\text{th}}}}{\text{ term, }}r = 5,{\text{ }}\left( {\begin{array}{*{20}{c}} 8 \\ 5 \end{array}} \right),{\text{ (2x}}{{\text{)}}^3}{(3)^5}\)

correct working (may be seen in expansion)     (A1)

eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 8 \\ 5 \end{array}} \right){(2x)^3}{(3)^5},{\text{ }}56 \times {2^3} \times {3^5}\)

\(108864{x^3}\;\;\;\)(accept \(109000{x^3}\))     A1     N3

[4 marks]

Notes:     Do not award any marks if there is clear evidence of adding instead of multiplying.

Do not award final A1 for a final answer of \(108864\), even if \(108864{x^3}\) is seen previously.

If no working shown award N2 for \(108864\).

This is a common question and yet it was not unusual to see candidates writing out the expansion in full or using Pascal’s triangle to find the correct binomial coefficient. Of those candidates who managed to identify the correct term, many omitted the parentheses around \(2\chi \) which led to an incorrect answer. Most candidates were able to distinguish between “the term in \({x^{3n}}\)” and the coefficient. There are still a significant number of candidates who add the parts of a term rather than multiply them and this approach gained no marks.

This is a common question and yet it was not unusual to see candidates writing out the expansion in full or using Pascal’s triangle to find the correct binomial coefficient. Of those candidates who managed to identify the correct term, many omitted the parentheses around \(2\chi \) which led to an incorrect answer. Most candidates were able to distinguish between “the term in \({x^{3n}}\)” and the coefficient. There are still a significant number of candidates who add the parts of a term rather than multiply them and this approach gained no marks.