IB Math Analysis & Approaches Questionbank-Topic: SL 1.9-The binomial theorem -SL Paper 1

Question

The fifth term in the expansion of the binomial \({(a + b)^n}\) is given by \(\left( {\begin{array}{*{20}{c}}
{10}\\
4
\end{array}} \right){p^6}{(2q)^4}\) .

Write down the value of \(n\).

[1]
a.

Write down a and b, in terms of p and/or q.

[2]
b.

Write down an expression for the sixth term in the expansion.

[3]
c.
Answer/Explanation

Markscheme

\(n = 10\)     A1     N1

[1 mark]

a.

\(a = p\) , \(b = 2q\) (or \(a = 2q\) , \(b = p\) )     A1A1     N1N1

[2 marks]

b.

\(\left( {\begin{array}{*{20}{c}}
{10}\\
5
\end{array}} \right){p^5}{(2q)^5}\)     A1A1A1     N3

[3 marks]

c.

Question

Expand \({(2 + x)^4}\) and simplify your result.

[3]
a.

Hence, find the term in \({x^2}\) in \({(2 + x)^4}\left( {1 + \frac{1}{{{x^2}}}} \right)\) .

[3]
b.
Answer/Explanation

Markscheme

evidence of expanding     M1

e.g. \({2^4} + 4({2^3})x + 6({2^2}){x^2} + 4(2){x^3} + {x^4}\) , \((4 + 4x + {x^2})(4 + 4x + {x^2})\)

\({(2 + x)^4} = 16 + 32x + 24{x^2} + 8{x^3} + {x^4}\)     A2     N2

[3 marks]

a.

finding coefficients 24 and 1     (A1)(A1)

term is \(25{x^2}\)     A1     N3

[3 marks]

b.

Question

Given that \({\left( {1 + \frac{2}{3}x} \right)^n}{(3 + nx)^2} = 9 + 84x + \ldots \) , find the value of n .

Answer/Explanation

Markscheme

attempt to expand \({\left( {1 + \frac{2}{3}x} \right)^n}\)     (M1)

e.g. Pascal’s triangle, \({\left( {1 + \frac{2}{3}x} \right)^n} = 1 + \frac{2}{3}nx + \ldots \)

correct first two terms of \({\left( {1 + \frac{2}{3}x} \right)^n}\) (seen anywhere)     (A1)

e.g. \(1 + \frac{2}{3}nx\)

correct first two terms of quadratic (seen anywhere)     (A1)

e.g. 9 , \(6nx\) , \((9 + 6nx + {n^2}{x^2})\)

correct calculation for the x-term     A2

e.g. \(\frac{2}{3}nx \times 9 + 6nx\) , \(6n + 6n\) , \(12n\)

correct equation     A1

e.g. \(6n + 6n = 84\) , \(12nx = 84x\)

\(n = 7\)     A1     N1

[7 marks]

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.

N16/5/MATME/SP1/ENG/TZ0/03

Write down the values in the fifth row of Pascal’s triangle.

[2]
a.

Hence or otherwise, find the term in \({x^3}\) in the expansion of \({(2x + 3)^5}\).

[5]
b.
Answer/Explanation

Markscheme

1, 5, 10, 10, 5, 1     A2     N2

[2 marks]

a.

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]

b.

Leave a Reply