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Question
In the expansion of \({(3x + 1)^n}\), the coefficient of the term in \({x^2}\) is \(135n\), where \(n \in {\mathbb{Z}^ + }\). Find \(n\).
Answer/Explanation
Markscheme
Note: Accept sloppy notation (such as missing brackets, or binomial coefficient which includes \({x^2}\)).
evidence of valid binomial expansion with binomial coefficients (M1)
eg\(\;\;\;\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){(3x)^r}{(1)^{n – r}},{\text{ }}{(3x)^n} + n{(3x)^{n – 1}} + \left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){(3x)^{n – 2}} + \ldots ,{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){(1)^{n – r}}{(3x)^r}\)
attempt to identify correct term (M1)
eg\(\;\;\;\left( {\begin{array}{*{20}{c}} n \\ {n – 2} \end{array}} \right),{\text{ }}{(3x)^2},{\text{ }}n – r = 2\)
setting correct coefficient or term equal to \(135n\) (may be seen later) A1
eg\(\;\;\;9\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right) = 135n,{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ {n – 2} \end{array}} \right){(3x)^2} = 135n,{\text{ }}\frac{{9n(n – 1)}}{2} = 135n{x^2}\)
correct working for binomial coefficient (using \(_n{C_r}\) formula) (A1)
eg\(\;\;\;\frac{{n(n – 1)(n – 2)(n – 3) \ldots }}{{2 \times 1 \times (n – 2)(n – 3)(n – 4) \ldots }},{\text{ }}\frac{{n(n – 1)}}{2}\)
EITHER
evidence of correct working (with linear equation in \(n\)) (A1)
eg\(\;\;\;\frac{{9(n – 1)}}{2} = 135,{\text{ }}\frac{{9(n – 1)}}{2}{x^2} = 135{x^2}\)
correct simplification (A1)
eg\(\;\;\;n – 1 = \frac{{135 \times 2}}{9},{\text{ }}\frac{{(n – 1)}}{2} = 15\)
\(n = 31\) A1 N2
OR
evidence of correct working (with quadratic equation in \(n\)) (A1)
eg\(\;\;\;9{n^2} – 279n = 0,{\text{ }}{n^2} – n = 30n,{\text{ (9}}{{\text{n}}^2} – 9n){x^2} = 270n{x^2}\)
evidence of solving (A1)
eg\(\;\;\;9n(n – 31) = 0,{\text{ }}9{n^2} = 279n\)
\(n = 31\) A1 N2
Note: Award A0 for additional answers.
[7 marks]
Question
The values in the fourth row of Pascal’s triangle are shown in the following table.
Write down the values in the fifth row of Pascal’s triangle.
Hence or otherwise, find the term in \({x^3}\) in the expansion of \({(2x + 3)^5}\).
Answer/Explanation
Markscheme
1, 5, 10, 10, 5, 1 A2 N2
[2 marks]
evidence of binomial expansion with binomial coefficient (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){a^{n – r}}{b^r}\), selecting correct term, \({(2x)^5}{(3)^0} + 5{(2x)^4}{(3)^1} + 10{(2x)^3}{(3)^2} + \ldots \)
correct substitution into correct term (A1)(A1)(A1)
eg\(\,\,\,\,\,\)\(10{(2)^3}{(3)^2},{\text{ }}\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right){(2x)^3}{(3)^2}\)
Note: Award A1 for each factor.
\(720{x^3}\) A1 N2
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 720, even if \(720{x^3}\) is seen previously.
[5 marks]