# IB Math Analysis & Approaches Question bank -Topic: SL 1.7 Laws of exponents- SL Paper 1

## Question

FindÂ $${\log _2}32$$ .

[1]
a.

Given thatÂ $${\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right)$$ can be written as $$px + qy$$ , find the value of p and of q.

[4]
b.

## Markscheme

5Â Â Â Â  A1 Â  Â  N1

[1 mark]

a.

METHOD 1

$${\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right) = {\log _2}{32^x} – {\log _2}{8^y}$$Â Â Â Â  (A1)

$$= x{\log _2}32 – y{\log _2}8$$Â Â Â Â  (A1)

$${\log _2}8 = 3$$Â Â Â Â  (A1)

$$p = 5$$ , $$q = – 3$$Â (accept $$5x – 3y$$Â )Â Â Â Â  A1Â Â Â Â Â  N3Â

METHOD 2

$$\frac{{{{32}^x}}}{{{8^y}}} = \frac{{{{({2^5})}^x}}}{{{{({2^3})}^y}}}$$Â Â Â Â  (A1)Â

$$= \frac{{{2^5}^x}}{{{2^3}^y}}$$Â Â Â Â  (A1)

$$= {2^{5x – 3y}}$$Â Â Â Â  (A1)

$${\log _2}({2^{5x – 3y}}) = 5x – 3y$$

$$p = 5$$ , $$q = – 3$$Â (accept $$5x – 3y$$Â )Â Â Â  Â A1Â Â Â Â Â  N3

[4 marks]

b.

## Question

Solve $${\log _2}x + {\log _2}(x – 2) = 3$$ , for $$x > 2$$ .

## Markscheme

recognizing $$\log a + \log b = \log ab$$ (seen anywhere)Â Â Â Â  (A1)

e.g. $${\log _2}(x(x – 2))$$ , $${x^2} – 2x$$

recognizing $${\log _a}b = x \Leftrightarrow {a^x} = b$$Â Â Â Â  (A1)

e.g. $${2^3} = 8$$

correct simplificationÂ Â Â Â  A1

e.g. $$x(x – 2) = {2^3}$$ , $${x^2} – 2x – 8$$

evidence of correct approach to solveÂ Â Â Â  (M1)

correct workingÂ Â Â Â  A1

e.g. $$(x – 4)(x + 2)$$ , $$\frac{{2 \pm \sqrt {36} }}{2}$$

$$x = 4$$Â Â Â Â  A2Â Â Â Â  N3

[7 marks]

Â

## Question

Write down the value of

(i) Â  Â  $${\log _3}27$$;

[1]
a(i).

(ii) Â  Â  $${\log _8}\frac{1}{8}$$;

[1]
a(ii).

(iii) Â  Â  $${\log _{16}}4$$.

[1]
a(iii).

Hence, solve $${\log _3}27 + {\log _8}\frac{1}{8} – {\log _{16}}4 = {\log _4}x$$.

[3]
b.

## Markscheme

(i) Â  Â  $${\log _3}27 = 3$$ Â  Â Â A1 Â  Â  N1

[1 mark]

a(i).

(ii) Â  Â  $${\log _8}\frac{1}{8} = Â – 1$$ Â  Â Â A1 Â  Â  N1

[1 mark]

a(ii).

(iii) Â  Â  $${\log _{16}}4 = \frac{1}{2}$$ Â  Â Â A1 Â  Â  N1

[1 mark]

a(iii).

correct equation with their three values Â  Â  (A1)

egÂ  Â  Â $$\frac{3}{2} = {\log _4}x{\text{, }}3 + ( – 1) – \frac{1}{2} = {\log _4}x$$

correct working involving powers Â  Â  (A1)

egÂ  Â  Â $$x = {4^{\frac{3}{2}}}{\text{, }}{4^{\frac{3}{2}}} = {4^{{{\log }_4}x}}$$

$$x = 8$$ Â  Â Â A1 Â  Â  N2

[3 marks]

b.

## Question

Find the value of each of the following, giving your answer as an integer.

$${\log _6}36$$

[2]
a.

$${\log _6}4 + {\log _6}9$$

[2]
b.

$${\log _6}2 – {\log _6}12$$

[3]
c.

## Markscheme

correct approach Â  Â  (A1)

eg Â  Â  $${6^x} = 36,{\text{ }}{6^2}$$

$$2$$ Â  Â  Â A1 Â  Â  N2

[2 marks]

a.

correct simplification Â  Â  (A1)

eg Â  Â  $${\log _6}36,{\text{ }}\log (4 \times 9)$$

$$2$$ Â  Â  Â A1Â  Â  Â  N2

[2 marks]

b.

correct simplification Â  Â  (A1)

eg Â  Â  $${\log _6}\frac{2}{{12}},{\text{ }}\log (2 \div 12)$$

correct working Â  Â  (A1)

eg Â  Â  $${\log _6}\frac{1}{6},{\text{ }}{6^{ – 1}} = \frac{1}{6}{,6^x} = \frac{1}{6}$$

$$-1$$ Â  Â Â A1 Â  Â  N2

[3 marks]

c.

## Question

Let $$x = \ln 3$$ and $$y = \ln 5$$. Write the following expressions in terms of $$x$$ and $$y$$.

$$\ln \left( {\frac{5}{3}} \right)$$.

[2]
a.

$$\ln 45$$.

[4]
b.

## Markscheme

correct approach Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\ln 5 – \ln 3$$

$$\ln \left( {\frac{5}{3}} \right) = y – x$$ Â  Â A1 Â  Â  N2

[2 marks]

a.

recognizing factors of 45 (may be seen in log expansion) Â  Â  (M1)

eg$$\,\,\,\,\,$$$$\ln (9 \times 5),{\text{ }}3 \times 3 \times 5,{\text{ }}\log {3^2} \times \log 5$$

correct application of $$\log (ab) = \log a + \log b$$Â Â  Â  (A1)

eg$$\,\,\,\,\,$$$$\ln 9 + \ln 5,{\text{ }}\ln 3 + \ln 3 + \ln 5,{\text{ }}\ln {3^2} + \ln 5$$

correct working Â  Â  (A1)

eg$$\,\,\,\,\,$$$$2\ln 3 + \ln 5,{\text{ }}x + x + y$$

$$\ln 45 = 2x + y$$ Â  Â A1 Â  Â  N3

[4 marks]

b.

## Question

An arithmetic sequence hasÂ $${u_1} = {\text{lo}}{{\text{g}}_c}\left( p \right)$$ and $${u_2} = {\text{lo}}{{\text{g}}_c}\left( {pq} \right)$$, whereÂ $$c > 1$$ and $$p,\,\,q > 0$$.

Show thatÂ $$d = {\text{lo}}{{\text{g}}_c}\left( q \right)$$.

[2]
a.

LetÂ $$p = {c^2}$$ andÂ $$q = {c^3}$$. Find the value of $$\sum\limits_{n = 1}^{20} {{u_n}}$$.

[6]
b.

## Markscheme

valid approach involving addition or subtractionÂ  Â  Â  Â M1
egÂ Â $${u_2} = {\text{lo}}{{\text{g}}_c}\,p + d,\,\,{u_1} – {u_2}$$

correct application of log lawÂ  Â  Â  A1
egÂ Â $${\text{lo}}{{\text{g}}_c}\left( {pq} \right) = {\text{lo}}{{\text{g}}_c}\,p + {\text{lo}}{{\text{g}}_c}\,q,\,\,{\text{lo}}{{\text{g}}_c}\left( {\frac{{pq}}{p}} \right)$$

$$d = {\text{lo}}{{\text{g}}_c}\,q$$Â Â  Â AG N0

[2 marks]

a.

METHOD 1 (findingÂ $${u_1}$$ and d)

recognizingÂ $$\sum { = {S_{20}}}$$Â (seen anywhere)Â  Â  Â  (A1)

attempt to findÂ $${u_1}$$ or d usingÂ $${\text{lo}}{{\text{g}}_c}\,{c^k} = k$$Â  Â  Â (M1)
egÂ Â $${\text{lo}}{{\text{g}}_c}\,c$$,Â $${\text{3}}\,{\text{lo}}{{\text{g}}_c}\,c$$, correct value ofÂ $${u_1}$$ orÂ d

$${u_1}$$ = 2,Â d = 3Â (seen anywhere)Â  Â  Â  (A1)(A1)

correct workingÂ  Â  Â (A1)
egÂ Â $${S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)$$

$$\sum\limits_{n = 1}^{20} {{u_n}}$$ = 610Â  Â  Â A1 N2

METHOD 2 (expressing S in terms of c)

recognizingÂ $$\sum { = {S_{20}}}$$Â (seen anywhere)Â  Â  Â Â (A1)

correct expression for S in terms of cÂ  Â  Â  (A1)
egÂ Â $$10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)$$

$${\text{lo}}{{\text{g}}_c}\,{c^2} = 2,\,\,\,{\text{lo}}{{\text{g}}_c}\,{c^3} = 3$$Â Â (seen anywhere)Â  Â  Â (A1)(A1)

correct workingÂ  Â  Â  (A1)

egÂ Â $${S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)$$

$$\sum\limits_{n = 1}^{20} {{u_n}}$$ = 610Â  Â  Â A1 N2

METHOD 3Â (expressingÂ SÂ in terms ofÂ c)

recognizingÂ $$\sum { = {S_{20}}}$$Â (seen anywhere)Â  Â  Â Â (A1)

correct expression forÂ SÂ in terms ofÂ cÂ  Â  Â Â (A1)
egÂ Â $$10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)$$

correct application of log lawÂ  Â  Â (A1)
egÂ Â $$2\,{\text{lo}}{{\text{g}}_c}\,{c^2} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^4},\,\,19\,{\text{lo}}{{\text{g}}_c}\,{c^3} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}},\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^2}} \right)}^2} + \,\,{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^3}} \right)}^{19}}} \right),\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{c^4} + \,{\text{lo}}{{\text{g}}_c}\,{c^{57}}} \right),\,\,10\left( {{\text{lo}}{{\text{g}}_c}\,{c^{61}}} \right)$$

correct application of definition of logÂ  Â  Â  (A1)
egÂ Â $${\text{lo}}{{\text{g}}_c}\,{c^{61}} = 61,\,\,{\text{lo}}{{\text{g}}_c}\,{c^4} = 4,\,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}} = 57$$

correct workingÂ  Â  Â (A1)
egÂ Â $${S_{20}} = \frac{{20}}{2}\left( {4 + 57} \right),\,\,10\left( {61} \right)$$

$$\sum\limits_{n = 1}^{20} {{u_n}}$$ = 610Â  Â  Â A1 N2

[6 marks]

b.

MAA SL 1.7 BINOMIAL THEOREM [concise]-lala

### Question

[Maximum mark: 9] [without GDC]
Write down the expansions of

(a) $$(1\pm x)^{3}$$

(b)Â  $$(1\pm x)^{4}$$

(c)Â  $$(1\pm x)^{5}$$

Ans.

(a) $$(1\pm x)^{3}=1\pm 3x+3x^{2}\pm x^{3}$$

(b) $$(1\pm x)^{4}=1\pm 4x+6x^{2}\pm 4x^{3}+x^{4}$$

(c) $$(1\pm x)^{5}=1\pm 5x+10x^{2}\pm 10x^{3}+5x^{4}\pm x^{5}$$

### Question

[Maximum mark: 10] [with GDC]
Write down the first four terms in ascending powers of x

(a) in the expansion of $$(1+x)^{10}$$

(b) in the expansion of $$(1+2x)^{10}$$

(c) in the expansion of $$(2-x)^{10}$$

Ans.

(a) $$(1+x)^{10}=1+10x+45x^{2}+120x^{3}+…$$

(b) $$(1+2x)^{10}=1+20x+180x^{2}+960x^{3}+…$$

(c) $$(2-x)^{10}=1024-5120x+11520x^{2}-15360x^{3}+…$$

### Question

[Maximum mark: 5] [with GDC]

Find the term inÂ  x7 in the expansion of $$(1-x)^{10}$$

Ans.

-120x7

### Question

[Maximum mark: 5] [with GDC]

Find the term inÂ  x16 in the expansion of $$(1-x^{2})^{10}$$

Ans.

45x16

### Question

[Maximum mark: 10] [with GDC]

(a) ExpandÂ  Â  Â  (i) $$\left ( x-\frac{1}{x} \right )^{3}$$Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (ii) $$\left ( x-\frac{1}{x} \right )^{4}$$

(b) In the expansion ofÂ  $$\left ( x-\frac{1}{x} \right )^{10}$$

(i) find the constant term
(ii) find the term in x2
(iii) find the term in x-2

Ans.

(a) (i) $$\left ( x-\frac{1}{x} \right )^{3}=x^{3}-3x+\frac{3}{x}-\frac{1}{x^{3}}$$Â  Â  Â  Â  Â  Â  Â  Â  (ii)Â  $$\left ( x-\frac{1}{x} \right )^{4}=x^{4}-4x^{2}+6-\frac{4}{x^{2}}-\frac{1}{x^{4}}$$

(b) (i) $$-\binom{10}{5}=-252$$Â  Â  Â  Â  Â  Â (ii) $$\binom{10}{4}=210$$Â  Â  Â  Â  Â  Â  Â (iii) $$\binom{10}{6}=210$$

Â

### Question

[Maximum mark: 10] [with GDC]

(a) ExpandÂ  $$(2x+1)^{4}$$

Hence, find
(b) the term in x2 in the expansion of $$x(2x+1)^{4}$$

(c) the term in x2 in the expansion of $$(x+1)(2x+1)^{4}$$

(d) the term in x2 in the expansion of $$(3-x^{2})(2x+1)^{4}$$

Ans.

(a) $$(1+2x)^{4}=1+8x+24x^{2}+32x^{3}+16x^{4}$$Â  Â  Â  Â  Â  Â  Â  Â  (b)Â  8x2Â  Â  Â  Â  Â  Â  Â (c) 32x2Â  Â  Â  Â  Â  Â  (d) 71x2

### Question

[Maximum mark: 10] [without GDC]
(a) Show that
(i) $$\binom{8}{7}=8$$,Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (ii) $$\binom{8}{6}=28$$,Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (iii) $$\binom{8}{5}=56$$
(b) Find
(i) $$\binom{10}{9}$$,Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (ii) $$\binom{10}{8}$$,Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (iii) $$\binom{10}{7}$$

Ans.

(a) (show)
(b)Â  Â  Â  Â  (i) 10Â  Â  Â  Â  Â  (ii) 45Â  Â  Â  Â  Â  Â (iii) 120

### Question

[Maximum mark: 8] [without GDC]

(a) Show that

(i) $$\binom{n}{0}=1$$,Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (ii) $$\binom{n}{1}=n$$,Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (iii) $$\binom{n}{2}=\frac{n(n-1)}{2}$$

(b) Simplify $$\binom{n}{3}$$

Ans.

(a) (show)

(b) $$\binom{n}{3}=\frac{n(n-1)(n-2)}{6}$$

### Question

[Maximum mark: 8] [without GDC]

(a) Verify that

$$\binom{5}{2}+\binom{5}{3}=\binom{6}{3}$$

(b) Prove that

$$\binom{19}{9}+\binom{19}{10}=\binom{20}{10}$$

Ans.

### Â

(a) (verify)

(b) LHS= $$\binom{19}{9}+\binom{19}{10}=\frac{19!}{9!x10!}+\frac{19!}{9!x10!}=\frac{2×19!}{9!x10!}=\frac{2×19!}{9!x10!}$$

RHS= $$\binom{20}{10}=\frac{20!}{10!x10!}=\frac{20×19!}{10×9!x10!}=\frac{2×19!}{9!x10!}$$

### Question

[Maximum mark: 6] [with / without GDC]

Complete the following expansion.

$$(2+ax)^{4}=16+32ax+…$$

Ans.

$$…+6×2^{2}(ax)^{2}+4×2(ax)^{3}+(ax)^{4}=…+24a^{2}x^{2}+8a^{3}x^{3}+a^{4}x^{4}$$

### Question

[Maximum mark: 4] [with GDC]

Use the binomial theorem to complete this expansion.

$$(3x+2y)^{4}=81x^{4}+216x^{3}y+…$$

Ans.

$$(3x+2y)^{4}=(3x)^{4}+\binom{4}{1}(3x)^{2}(2y)+\binom{4}{2}(3x)^{2}(2y)^{2}+\binom{4}{3}(3x)(2y)^{3}+(2y)^{4}$$
=$$81x^{4}+216x^{3}y+216x^{2}y^{2}+96xy^{3}+16y^{4}$$

### Question

[Maximum mark: 6] [with / without GDC]

Given that $$(3+\sqrt{7})^{3}=p+q\sqrt{7}$$ where $$p$$ and $$q$$ are integers, find

(a) $$p$$;Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (b) $$q$$.

Ans.

METHOD 1

Using binomial expansion

$$(3+\sqrt{7})^{3}=3^{3}+\binom{3}{1}3^{2}(\sqrt{7})+\binom{3}{2}3(\sqrt{7})^{2}+(\sqrt{7})^{3}=27+27\sqrt{7}+63+7\sqrt{7}$$

$$(3+\sqrt{7})^{3}=90+34\sqrt{7}$$Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (so p=90, q=34)

Â

METHOD 2

For multiplying

$$(3+\sqrt{7})^{2}(3+\sqrt{7})=(9+6\sqrt{7}+7)(3+\sqrt{7})=27+9\sqrt{7}+18\sqrt{7}+42+21+7\sqrt{7}$$

$$(3+\sqrt{7})^{3}=90+34\sqrt{7}$$Â  Â  Â  Â  Â  Â  Â  Â  Â  (so p=90, q=34)

### Question

[Maximum mark: 4] [with / without GDC]

Find the coefficient of a5b7 in the expansion of $$(a+b)^{12}$$.

Ans.

(a + b)12

Coefficient of a5b7 is $$\binom{12}{5}=\binom{12}{7}=792$$

### Question

[Maximum mark: 4] [with / without GDC]

Find the coefficient of a3b4 in the expansion of $$(5a+b)^{7}$$.

Ans.

$$(5a+b)^{7}=…+\binom{7}{4}(5a)^{3}(b)^{4}+…=\frac{7x6x5x4}{1x2x3x4}x5^{3}x(a^{3}b^{4})=35×5^{3}xa^{3}b^{4}$$

So the coefficient is 4375

### Question

[Maximum mark: 6] [without GDC]

The fifth term in the expansion of the binomial $$(a+b)^{n}$$ is given by $$\binom{10}{4}p^{6}(2q)^{4}$$.

(a) Write down the value of $$n$$.
(b) Write down $$a$$ and $$b$$, in terms of $$p$$ and/or $$q$$.
(c) Write down an expression for the sixth term in the expansion.

Ans.

(a) n = 10
(b) a = p, b = 2q (or a = 2q, b = p)

(c) $$\binom{10}{5}p^{5}(2q)^{5}$$

### Question

[Maximum mark: 5] [with / without GDC]

Consider the expansion of $$(x+2)^{11}.$$

(a) Write down the number of terms in this expansion.
[b) Find the term containing x2.

Ans.

(a) 12 terms

(b) (10th term, $$r=9, \binom{11}{9}(x)^{2}(2)^{9})$$
$$\binom{11}{9}(x)^{2}(2)^{9}=55×2^{9}x^{2}=28160x^{2}$$

### Question

[Maximum mark: 6] [with / without GDC]

Find the coefficient of x3 in the expansion of $$(2-x)^{5}$$.

Ans.

Term involving x3 isÂ  $$\binom{5}{3}(2)^{2}(-x)^{3}=-40x^{3}\Rightarrow$$ The coefficient is â€“40

### Question

[Maximum mark: 4] [with GDC]

Find the coefficient of x5 in the expansion of $$(3x-2)^{8}$$.

Ans.

Required term is $$\binom{8}{5}(3x)^{5}(-2)^{3}$$

Therefore the coefficient of x5 is 56 Ã— 243 Ã— â€“8 = â€“108864

### Question

[Maximum mark: 6] [with GDC]

Find the term containing x3 in the expansion of $$(2-3x)^{8}$$.

Ans.

$$\binom{8}{3}(2)^{5}(-3x)^{3}$$
Term is $$-48384x^{3}$$

### Question

[Maximum mark: 5] [with GDC]

Find the term in x3 in the expansion of $$\left ( \frac{2}{3}x-3 \right )^{8}$$.

Ans.

$$\binom{8}{5}\left ( \frac{2}{3}x \right )^{3}(-3)^{5}=-4032x^{3}$$

### Question

[Maximum mark: 6] [with / without GDC]

One of the terms of the expansion of $$(x+2y)^{10}$$ is $$ax^{8}y^{2}$$. Find the value of $$a$$.

Ans.

$$a=\binom{10}{8}x2^{2}=45×4=180$$

### Question

[Maximum mark: 6] [with GDC]

Find the term containing x10 in the expansion of $$(5+2x^{2})^{7}$$.

Ans.

$$\left ( \frac{7}{2} \right )5^{2}(2x^{2})^{5}=16\;800x^{10}$$

### Question

[Maximum mark: 6] [with GDC]

Consider the expansion ofÂ  $$(x^{2}-2)^{5}$$.

(a) Write down the number of terms in this expansion.
(b) The first four terms of the expansion in descending powers of $$x$$ are

$$x^{10}-10x^{8}+40x^{6}+Ax^{4}+…$$

Find the value of $$A$$.

Ans.

(a) 6 terms

(b) the fourth term is $$\binom{5}{3}(-2)^{3}(x^{2})^{2}=10(-8)x^{4}=-80x^{4}$$Â  Â  Â  Â hence $$A$$=-80

### Question

[Maximum mark: 6] [with GDC]

Consider the expansion of $$\left ( 3x^{2}-\frac{1}{x} \right )^{9}$$.

(a) How many terms are there in this expansion?
(b) Find the constant term in this expansion.

Ans.

(a) 10

(b)Â  $$\binom{9}{6}(3x^{2})^{3}\left ( -\frac{1}{x} \right )^{6}=84×3^{3}x^{6}\frac{1}{x^{6}}$$

constant =2268

### Question

[Maximum mark: 4] [with GDC]

Determine the constant term in the expansion of $$\left ( x-\frac{2}{x^{2}} \right )^{9}$$.

Ans.

$$\binom{9}{3}x^{6}\left ( \frac{-2}{x^{2}} \right )^{3}=84x^{6}\left ( \frac{-8}{x^{6}} \right )=-672$$

### Question

[Maximum mark: 6] [with GDC]

Consider the expansion of the expression $$(x^{3}-3x)^{6}$$.

(a) Write down the number of terms in this expansion.
(b) Find the term in x12.

Ans.

(a) 7 terms

(b) $$\binom{6}{3}(x^{3})^{3}(-3x)^{3}$$
$$\binom{6}{3}=20,(-3)^{3}=-27$$

Term is -540x12

### Question

[Maximum mark: 6] [with / without GDC]

Find the term in x4 in the expansion of $$\left ( 3x^{2}-\frac{2}{x} \right )^{5}$$.

Ans.

$$\binom{5}{2}(3x^{2})^{3}\left ( \frac{-2}{x} \right )^{2}\binom{5}{2},27x^{6},\frac{4}{x^{2}};10(3x^{2})^{3}\left ( \frac{-2}{x} \right )^{2}=10x27x4x^{4}$$

term = 1080x4

### Question

[Maximum mark: 6] [with GDC]

When the expression $$(2+ax)^{10}$$ is expanded, the coefficient of the term in x3 is 414 720.
Find the value of $$a$$.

Ans.

$$\binom{10}{3}2^{7}(ax)^{3}$$Â  Â  Â  Â  Â  Â  Â  Â  Â  $$\left ( accept\binom{10}{7} \right )$$
$$120×2^{7}a^{3}=414\;720$$
$$a^{3}=27\Leftrightarrow a=3$$

### Question

[Maximum mark: 6] [without GDC]

(a) Expand $$(x-2)^{4}$$ and simplify your result.

(b) Find the term in x3 in $$(3x+4)(x-2)^{4}$$.

Ans.

(a) $$(x-2)^{4}=x^{4}+4x^{3}(-2)+6x^{2}(-2)^{2}+4x(-2)^{3}+(-2)^{4}$$
$$(x-2)^{4}=x^{4}-8x^{3}+24x^{2}-32x+16$$

(b) finding coefficients, 3 Ã— 24 (= 72), 4 Ã— (â€“8)(= â€“32)
term is 40x3

### Question

[Maximum mark: 6] [without GDC]

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

(b) Hence, find the term in x2 in $$(2+x)^{4}\left ( 1+\frac{1}{x^{2}} \right )$$.

Ans.

(a)Â  $$2^{4}+4(2^{3})x+6(2^{2})x^{2}+4(2)x^{3}+x^{4},(4+4x+x^{2})(4+4x+x^{2})$$
$$(2+x)^{4}=16+32x+24x^{2}+8x^{3}+x^{4}$$

(b) finding coefficients 24 and 1
term is 25x2

A

### Question

[Maximum mark: [4] [without GDC]

Consider the binomial expansion $$(1+x)^{4}=1+\binom{4}{1}x+\binom{4}{2}x^{2}+\binom{4}{3}x^{3}+x^{4}$$

(a) By substituting x = 1 into both sides, or otherwise, evaluate $$\binom{4}{1}+\binom{4}{2}+\binom{4}{3}$$.

(b) Evaluate $$\binom{9}{1}+\binom{9}{2}+\binom{9}{3}+\binom{9}{4}+\binom{9}{5}+\binom{9}{6}+\binom{9}{7}+\binom{9}{8}$$.

Extra question

Find a similar result by substituting x =- 1 in the expansion of (1+ x)4 above.

Ans.

(a)Â  $$(1+1)^{4}=2^{4}=1+\binom{4}{1}(1)+\binom{4}{2}(1^{2})+\binom{4}{3}(1^{3})+1^{4}$$
$$\Rightarrow \binom{4}{1}+\binom{4}{2}+\binom{4}{3}=16-2=14$$

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(b)Â  Â $$(1+1)^{9}=1+\binom{9}{1}+\binom{9}{2}+\binom{9}{3}+…+\binom{9}{8}+1$$
$$\Rightarrow \binom{9}{1}+\binom{9}{2}+\binom{9}{3}+…+\binom{9}{8}=2^{9}-2=510$$

Extra question

$$0=1-\binom{4}{1}+\binom{4}{2}-\binom{4}{3}+1\Rightarrow \binom{4}{1}-\binom{4}{2}+\binom{4}{3}=2$$

### Question

[Maximum mark: 6] [without GDC]

(a) Expand $$\left ( e+\frac{1}{e} \right )^{4}$$ in terms of e.

(b) Express $$\left ( e+\frac{1}{e} \right )^{4}+\left ( e-\frac{1}{e} \right )^{4}$$ as the sum of three terms.

(a)Â  $$\left ( e+\frac{1}{e} \right )^{4}=\binom{4}{0}e^{4}+\binom{4}{1}e^{3}\left ( \frac{1}{e} \right )+\binom{4}{2}e^{2}\left ( \frac{1}{e} \right )^{2}+\binom{4}{3}e\left ( \frac{1}{e} \right )^{3}+\binom{4}{4}\left ( \frac{1}{e} \right )^{4}$$
$$\left ( e+\frac{1}{e} \right )^{4}=e^{4}+4e^{3}\left ( \frac{1}{e} \right )+6e^{2}\left ( \frac{1}{e} \right )^{2}+4e\left ( \frac{1}{e} \right )^{3}+\left ( \frac{1}{e} \right )^{4}=e^{4}+4e^{2}+6+\frac{4}{e^{2}}+\frac{1}{e^{4}}$$
(b) $$\left ( e-\frac{1}{e} \right )^{4}=e^{4}-4e^{3}\left ( \frac{1}{e} \right )+6e^{2}\left ( \frac{1}{e} \right )^{2}-4e\left ( \frac{1}{e} \right )^{3}+\left ( \frac{1}{e} \right )^{4}=e^{4}-4e^{2}+6-\frac{4}{e^{2}}+\frac{1}{e^{4}}$$
Adding gives $$2e^{4}+12+\frac{2}{e^{4}}$$