Consider the expansion of \({\left( {2x + \frac{k}{x}} \right)^9}\), where k > 0 . The coefficient of the term in x³ is equal to the coefficient of the term in  x⁵. Find k.

valid approach to find one of the required terms (must have correct substitution for parameters but accept “r” or an incorrect value for r)       (M1)
eg    \(\left( \begin{gathered}
9 \hfill \\
r \hfill \\
\end{gathered} \right){\left( {2x} \right)^{9 - r}}{\left( {\frac{k}{x}} \right)^r},\,\,\left( \begin{gathered}
9 \hfill \\
6 \hfill \\
\end{gathered} \right){\left( {2x} \right)^6}{\left( {\frac{k}{x}} \right)^3},\,\,\left( \begin{gathered}
9 \hfill \\
0 \hfill \\
\end{gathered} \right){\left( {2x} \right)^0}{\left( {\frac{k}{x}} \right)^9} + \left( \begin{gathered}
9 \hfill \\
1 \hfill \\
\end{gathered} \right){\left( {2x} \right)^1}{\left( {\frac{k}{x}} \right)^8} + \)…, Pascal’s triangle to 9th row

Note: Award M0 if there is clear evidence of adding instead of multiplying.

identifying correct terms (must be clearly indicated if only seen in expansion)      (A1)(A1)

eg  for x³ term: r = 3, r = 6, 7th term, \(\left( \begin{gathered}
9 \hfill \\
6 \hfill \\
\end{gathered} \right),\,\,\left( \begin{gathered}
9 \hfill \\
3 \hfill \\
\end{gathered} \right),\,\,{\left( {2x} \right)^6}{\left( {\frac{k}{x}} \right)^3},\,\,5376{k^3}\)

for x⁵ term: r = 2, r = 7, 8th term, \(\left( \begin{gathered}
9 \hfill \\
7 \hfill \\
\end{gathered} \right),\,\,\left( \begin{gathered}
9 \hfill \\
2 \hfill \\
\end{gathered} \right),\,\,{\left( {2x} \right)^7}{\left( {\frac{k}{x}} \right)^2},\,\,4608{k^2}\)

correct equation (may include powers of x)       A1
eg  \(\left( \begin{gathered}
9 \hfill \\
3 \hfill \\
\end{gathered} \right){\left( {2x} \right)^6}{\left( {\frac{k}{x}} \right)^3} = \left( \begin{gathered}
9 \hfill \\
2 \hfill \\
\end{gathered} \right){\left( {2x} \right)^7}{\left( {\frac{k}{x}} \right)^2}\)

valid attempt to solve their equation in terms of k only      (M1)
eg  sketch, \(84 \times 64{k^3} - 36 \times 128{k^2} = 0,\,\,5376k - 4608 = 0,\,\,\left( \begin{gathered}
9 \hfill \\
3 \hfill \\
\end{gathered} \right){2^6}{k^3} = \left( \begin{gathered}
9 \hfill \\
2 \hfill \\
\end{gathered} \right){2^7}{k^2}\)

0.857142

\(k = \frac{{4608}}{{5376}}\left( { = \frac{6}{7}} \right)\) (exact), 0.857     A1N4

[6 marks]