Free worked solutions · 2026 Edition

IB Maths AA HL Chapter 3: Binomial theorem worked solutions

Worked solutions to the practice questions in Chapter 3 of Photon Guidebook: Mathematics Analysis and Approaches HL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[5 marks]

(a) Use Pascal's triangle to expand (p+q)^5.

[2]

(b) Expand and simplify (x-1)^4.

[3]

Show solutionHide solution

(a) coefficients 1,\ 5,\ 10,\ 10,\ 5,\ 1

p^5+5p^4q+10p^3q^2+10p^2q^3+5pq^4+q^5

(b) use of row 1,\ 4,\ 6,\ 4,\ 1 with b=-1

x^4-4x^3+6x^2-4x+1

Note: award A1 for one sign or coefficient error.

Final answer

(a) p^5+5p^4q+10p^3q^2+10p^2q^3+5pq^4+q^5
(b) x^4-4x^3+6x^2-4x+1

Question 2

[5 marks]

The general term in the expansion of (a+b)^n is \binom{n}{r}a^{n-r}b^r, for r=0,1,\dots,n.

Consider the expansion of (x+3)^7.

(a) Write down the number of terms in the expansion.

[1]

(b) Write down the general term of the expansion.

[1]

(c) Find the term in x^5.

[3]

Show solutionHide solution

(a) 8

(b) \binom{7}{r}x^{7-r}3^r

(c) 7-r=5 \Rightarrow r=2 (A1)

\binom{7}{2}x^5(3)^2=21\times 9\,x^5

=189x^5

Final answer

(a) 8
(b) \binom7r x^{7-r}3^r
(c) 189x^5

Question 3

[6 marks]

Consider the expansion of \left(x+\dfrac{3}{x}\right)^6.

(a) Show that the general term can be written as \binom{6}{r}3^r x^{6-2r}.

[2]

(b) Find the constant term.

[2]

(c) Find the term in \dfrac{1}{x^2}.

[2]

Show solutionHide solution

(a) \binom{6}{r}x^{6-r}\left(\dfrac{3}{x}\right)^r=\binom{6}{r}x^{6-r}\,3^r x^{-r}

=\binom{6}{r}3^r x^{6-2r}

A1AG

(b) 6-2r=0 \Rightarrow r=3

\binom{6}{3}3^3=20\times 27=540

(c) 6-2r=-2 \Rightarrow r=4

\binom{6}{4}3^4x^{-2}=\dfrac{1215}{x^2}

Final answer

(a) shown
(b) 540
(c) \dfrac{1215}{x^2}

Question 4

[5 marks]

(a) Find the first three terms, in ascending powers of x, of the expansion of (1-x)^5.

[2]

(b) Hence find the coefficient of x^2 in the expansion of (3+2x)(1-x)^5.

[3]

Show solutionHide solution

(a) 1-5x+10x^2 A1A1

Note: A1 for -5x, A1 for 10x^2.

(b) x^2 terms come from 3\times 10x^2 and 2x\times(-5x)

30x^2-10x^2

coefficient =20

Final answer

(a) 1-5x+10x^2
(b) 20

Question 5

[5 marks]

The coefficient of x^2 in the expansion of (1+3x)^n, where n\ge 2 is an integer, is 189.

(a) Show that n^2-n-42=0.

[3]

(b) Hence find the value of n.

[2]

Show solutionHide solution

(a) \binom{n}{2}(3x)^2, coefficient \dfrac{n(n-1)}{2}\times 9 M1A1

\dfrac{9n(n-1)}{2}=189 \Rightarrow n(n-1)=42

n^2-n-42=0

(b) (n-7)(n+6)=0

n=7 (since n>0)

Final answer

(a) shown
(b) n=7

Question 6

[6 marks]

Do not use a calculator in this question.

(a) Write down the expansion of (a+b)^4.

[1]

(b) By writing 101=100+1, find the exact value of 101^4.

[3]

(c) Hence, or otherwise, find the exact value of 99^4.

[2]

Show solutionHide solution

(a) a^4+4a^3b+6a^2b^2+4ab^3+b^4

(b) 100^4+4(100)^3+6(100)^2+4(100)+1

=100\,000\,000+4\,000\,000+60\,000+400+1 (A1)

=104\,060\,401

(c) 99^4=(100-1)^4=100\,000\,000-4\,000\,000+60\,000-400+1

=96\,059\,601

Final answer

(a) a^4+4a^3b+6a^2b^2+4ab^3+b^4
(b) 104\,060\,401
(c) 96\,059\,601

Question 7

[6 marks]

For n \in \mathbb{Q} and \lvert x \rvert < 1,

\begin{gathered}(1+x)^{n} = 1 + nx + \frac{n(n-1)}{2!}x^{2}\\ {}+ \frac{n(n-1)(n-2)}{3!}x^{3} + \ldots\end{gathered}

(a) Find the first four terms in the expansion of (1+x)^{-3}.

[3]

(b) Find the first four terms in the expansion of (1+x)^{\frac{1}{3}}.

[3]

Show solutionHide solution

(a) 1 + (-3)x + \dfrac{(-3)(-4)}{2!}x^2 + \dfrac{(-3)(-4)(-5)}{3!}x^3

= 1 - 3x + 6x^2 - 10x^3 A1A1

(b) 1 + \dfrac{1}{3}x + \dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2!}x^2 + \dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}x^3

= 1 + \dfrac{1}{3}x - \dfrac{1}{9}x^2 + \dfrac{5}{81}x^3 A1A1

Note: in each part award A1A1 for all four terms correct, A1A0 for two or three correct terms.

Final answer

(a) 1-3x+6x^2-10x^3
(b) 1+\frac x3-\frac{x^2}{9}+\frac{5x^3}{81}

Question 8

[6 marks]

In the expansion of (1 + kx)^{\frac{3}{2}} in ascending powers of x, where k > 0, the coefficient of x^2 is \dfrac{147}{8}.

(a) Find the value of k.

[3]

(b) Find the coefficient of x^3.

[2]

(c) State the values of x for which the expansion is valid.

[1]

Show solutionHide solution

(a) \dfrac{\frac{3}{2} \times \frac{1}{2}}{2!}k^2 = \dfrac{3}{8}k^2

\dfrac{3}{8}k^2 = \dfrac{147}{8} \Rightarrow k^2 = 49

k = 7 (since k > 0)

(b) \dfrac{\frac{3}{2} \times \frac{1}{2} \times \left(-\frac{1}{2}\right)}{3!}(7x)^3 = -\dfrac{1}{16} \times 343x^3

coefficient = -\dfrac{343}{16}

(c) \lvert 7x \rvert < 1 \Rightarrow \lvert x \rvert < \dfrac{1}{7}

Final answer

(a) k=7
(b) -\dfrac{343}{16}
(c) |x|<\frac17

Question 9

[8 marks]

(a) Find the first three terms in the expansion of \sqrt{4 - x} in ascending powers of x.

[3]

(b) State the values of x for which this expansion is valid.

[1]

(c) Show that \sqrt{3.96} = \dfrac{3\sqrt{11}}{5}.

[1]

(d) Hence, by substituting a suitable value of x into your answer to (a), find an approximation for \sqrt{11}, giving your answer correct to 6 decimal places.

[3]

Show solutionHide solution

(a) \sqrt{4 - x} = 2\left(1 - \dfrac{x}{4}\right)^{\frac{1}{2}} = 2\left[1 + \dfrac{1}{2}\left(-\dfrac{x}{4}\right) + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2!}\left(-\dfrac{x}{4}\right)^2\right]

= 2\left[1 - \dfrac{x}{8} - \dfrac{x^2}{128}\right]

= 2 - \dfrac{x}{4} - \dfrac{x^2}{64}

(b) \left\lvert \dfrac{x}{4} \right\rvert < 1 \Rightarrow \lvert x \rvert < 4

(c) \sqrt{3.96} = \sqrt{\dfrac{396}{100}} = \dfrac{\sqrt{36 \times 11}}{10} = \dfrac{6\sqrt{11}}{10} = \dfrac{3\sqrt{11}}{5} A1AG

(d) x = 0.04 (inside \lvert x \rvert < 4): \sqrt{3.96} \approx 2 - 0.01 - 0.000025 = 1.989975 M1A1

\sqrt{11} \approx \dfrac{5}{3} \times 1.989975 = 3.316625

Note: the calculator value is 3.3166248\ldots, which also rounds to 3.316625: the approximation is correct to 6 decimal places.

Final answer

(a) 2-\frac x4-\frac{x^2}{64}
(b) |x|<4
(c) shown
(d) 3.316625

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 10

[5 marks]

In the expansion of (1+x)^n, where n>3, the coefficient of x^3 is five times the coefficient of x. Find the value of n.

Show solutionHide solution

\binom{n}{3}=5\binom{n}{1}

[1]

\dfrac{n(n-1)(n-2)}{6}=5n

[1]

Dividing by n\neq0: (n-1)(n-2)=30

[1]

n^2-3n-28=0\Rightarrow(n-7)(n+4)=0

[1]

n>3, so n=7 (reject n=-4).

[1]

Final answer

n=7

Question 11

[6 marks]

The coefficients of x^4, x^5 and x^6 in the expansion of (1+x)^n are three consecutive terms of an arithmetic sequence. Find the possible values of n.

Show solutionHide solution

For three consecutive terms of an arithmetic sequence, twice the middle equals the sum of the outer two:

[1]

2\binom{n}{5}=\binom{n}{4}+\binom{n}{6}.

[1]

Divide by \binom{n}{4}, using \binom{n}{5}\Big/\binom{n}{4}=\dfrac{n-4}{5} and \binom{n}{6}\Big/\binom{n}{4}=\dfrac{(n-4)(n-5)}{30}:

[1]

\dfrac{2(n-4)}{5}=1+\dfrac{(n-4)(n-5)}{30}

[1]

12(n-4)=30+(n-4)(n-5)\Rightarrow n^2-21n+98=0

[1]

(n-7)(n-14)=0, so n=7 or n=14 (both are at least 6, so both are valid).

[1]

Final answer

n=7 or n=14

Question 12

[7 marks]

Consider the expansion of (2+3x)(1+x)^n, where n\geq3.

(a) Show that the coefficient of x^3 is \dfrac{n(n-1)(2n+5)}{6}.

[4]

(b) The coefficient of x^3 is 196. Find the value of n.

[3]

Show solutionHide solution

(a) Pairs: 2\times\left(\text{coefficient of }x^3\right)+3\times\left(\text{coefficient of }x^2\right) in (1+x)^n.

[1]

2\binom{n}{3}+3\binom{n}{2}=\dfrac{n(n-1)(n-2)}{3}+\dfrac{3n(n-1)}{2}

[1]

=\dfrac{n(n-1)\left[2(n-2)+9\right]}{6}

[1]

=\dfrac{n(n-1)(2n+5)}{6}. [1]

(b) n(n-1)(2n+5)=1176.

[1]

n=8 works: 8\times7\times21=1176.

[1]

Factorising, (n-8)(2n^2+19n+147)=0, and 2n^2+19n+147 has discriminant 361-1176<0, so n=8 is the only solution.

[1]

Final answer

(a) shown
(b) n=8

Question 13

[10 marks]

(a) Show that (4+\sqrt{15})^4+(4-\sqrt{15})^4=3842.

[3]

(b) Explain why (4+\sqrt{15})^n+(4-\sqrt{15})^n is an integer for every n\in\mathbb{Z}^+.

[3]

(c) Show that 4-\sqrt{15}<\dfrac17, and hence show that (4+\sqrt{15})^{10} is within 10^{-8} of an integer.

[4]

Show solutionHide solution

(a) (4+\sqrt{15})^4=4^4+4(4^3)\sqrt{15}+6(16)(15)+4(4)(15\sqrt{15})+15^2

[1]

=256+256\sqrt{15}+1440+240\sqrt{15}+225=1921+496\sqrt{15}.

[1]

Replacing \sqrt{15} by -\sqrt{15} gives (4-\sqrt{15})^4=1921-496\sqrt{15}, so the sum is 3842.

[1]

(b) Expand both brackets by the binomial theorem; the terms are \binom{n}{r}4^{n-r}(\sqrt{15})^r and \binom{n}{r}4^{n-r}(-\sqrt{15})^r.

[1]

For odd r the two terms are equal and opposite, so all irrational (odd-r) terms cancel.

[1]

For even r the term is \binom{n}{r}4^{n-r}15^{r/2}, an integer, appearing twice; so the sum is an integer.

[1]

(c) \sqrt{15}>\dfrac{27}{7} because 15=\dfrac{735}{49}>\dfrac{729}{49}, so 4-\sqrt{15}<4-\dfrac{27}{7}=\dfrac17. [1]

[1]

Also 4-\sqrt{15}>0, so 0<(4-\sqrt{15})^{10}<\left(\dfrac17\right)^{10}=\dfrac{1}{282\,475\,249}<10^{-8}.

[1]

By part (b), (4+\sqrt{15})^{10}=N-(4-\sqrt{15})^{10} for an integer N, so it is within 10^{-8} of N.

[1]

Final answer

(a) shown
(b) the odd powers of \sqrt{15} cancel and the even powers are integers
(c) shown

Question 14

[5 marks]

State the range of values of x for which each expansion is valid.

(a) \left(1-\dfrac x4\right)^{-2}

[1]

(b) (2+5x)^{1/2}

[2]

(c) (1+3x)^7(1-2x)^{-1/2}

[2]

Show solutionHide solution

(a) \left|\dfrac x4\right|<1\Rightarrow|x|<4.

[1]

(b) (2+5x)^{1/2}=\sqrt2\left(1+\dfrac{5x}{2}\right)^{1/2}, valid when \left|\dfrac{5x}{2}\right|<1

[1]

|x|<\dfrac25.

[1]

(c) (1+3x)^7 is a finite polynomial, valid for every x, so it adds no restriction.

[1]

Only (1-2x)^{-1/2} restricts: |2x|<1\Rightarrow|x|<\dfrac12.

[1]

Final answer

(a) |x|<4
(b) |x|<\frac25
(c) |x|<\frac12

Question 15

[5 marks]

Find the first three terms, in ascending powers of x, of the expansion of \dfrac{1}{\sqrt{25-4x}}, and state the range of values of x for which the expansion is valid.

Show solutionHide solution

\dfrac{1}{\sqrt{25-4x}}=(25-4x)^{-1/2}=25^{-1/2}\left(1-\dfrac{4x}{25}\right)^{-1/2}=\dfrac15\left(1-\dfrac{4x}{25}\right)^{-1/2}. [1]

[1]

\left(1-\dfrac{4x}{25}\right)^{-1/2}=1-\dfrac12\left(-\dfrac{4x}{25}\right)+\dfrac{\left(-\frac12\right)\left(-\frac32\right)}{2}\left(\dfrac{4x}{25}\right)^2+\cdots=1+\dfrac{2x}{25}+\dfrac{6x^2}{625}+\cdots

[1]

\dfrac{1}{\sqrt{25-4x}}=\dfrac15+\dfrac{2x}{125}+\dfrac{6x^2}{3125}+\cdots

[1]

Valid when \left|\dfrac{4x}{25}\right|<1, that is |x|<\dfrac{25}{4}.

[1]

Final answer

\dfrac15+\dfrac{2x}{125}+\dfrac{6x^2}{3125}; |x|<\dfrac{25}{4}

Question 16

[6 marks]

Let Q(u)=(1+5u)^{2/3}/(1+ku), where k\in\mathbb{R}. The expansion of Q(u) in ascending powers of u has no term in u.

(a) Find k.

[3]

(b) Find the coefficient of u^{2}.

[2]

(c) State the values of u for which the expansion is valid.

[1]

Show solutionHide solution

(a) The numerator begins 1+10u/3-25u^2/9 and the reciprocal denominator begins 1-ku+k^2u^2. The linear coefficient is 10/3-k, giving k=10/3. (b) The quadratic coefficient is -25/9+k^2-(10/3)k=-25/9. (c) The respective radii are 1/5 and 3/10, so jointly |u|<1/5.

Final answer

(a) k=\frac{10}{3}
(b) -\frac{25}{9}
(c) |u|<\frac15

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 17

[4 marks]

The expression \left(x - \dfrac{2}{x}\right)^{8} is expanded. Determine the coefficient of the term \dfrac{1}{x^{2}}.

[4]

Show solutionHide solution

The general term is

\binom{8}{r}x^{8-r}\left(-\frac{2}{x}\right)^{r} = \binom{8}{r}(-2)^{r}x^{\,8-2r}

For x^{-2} we need 8 - 2r = -2, so r = 5:

\binom{8}{5}(-2)^{5} = 56 \times (-32) = \mathbf{-1792}

A1[4]

Final answer

-1792

Question 18

[5 marks]

In the expansion of \left(16x^{4} - \dfrac{1}{2x}\right)^{n}, where n \in \mathbb{Z}^{+}, determine every value of n for which a non-zero constant term occurs.

[5]

Show solutionHide solution

The general term of the expansion is

T_{r+1} = \binom{n}{r}\left(16x^{4}\right)^{n-r}\left(-\frac{1}{2x}\right)^{r}

Collecting the powers of x:

x^{4(n-r)} \times x^{-r} = x^{4n - 5r}

A constant term requires this exponent to vanish:

4n - 5r = 0 \implies r = \frac{4n}{5}

Since r must be an integer with 0 \le r \le n, the quantity \dfrac{4n}{5} must be a whole number, and as 4 and 5 share no factor this forces 5 \mid n. Any such r also satisfies r = \dfrac{4n}{5} \le n, and the coefficient \binom{n}{r}16^{n-r}\left(-\frac{1}{2}\right)^{r} is never zero.

n is any positive multiple of 5, i.e. n = 5,\ 10,\ 15,\ \ldots

[5]

Final answer

every positive multiple of 5

Question 19

[5 marks]

In the expansion of (x + k)^{8}, where k \in \mathbb{R}, the term in x^{6} has coefficient 175.

Determine the possible values of k.

[5]

Show solutionHide solution

In the binomial expansion of (x + k)^{8}, the term containing x^{6} requires two factors of k:

\binom{8}{2}x^{6}k^{2}, \qquad \binom{8}{2} = 28

Equating the coefficient to 175:

28k^{2} = 175 \implies k^{2} = \frac{25}{4}

k = \pm\frac{5}{2}

k = \dfrac{5}{2} or k = -\dfrac{5}{2}

[5]

Final answer

k=\pm\frac52

Question 20

[5 marks]

For a positive integer n \geq 4, the expansion of (2 + x)^{n} has equal coefficients of x^{3} and of x^{4}. Determine n.

[5]

Show solutionHide solution

The general term is \displaystyle\binom{n}{r}2^{\,n-r}x^{r}, so the coefficients of x^{3} and x^{4} are

\binom{n}{3}2^{\,n-3} \quad \text{and} \quad \binom{n}{4}2^{\,n-4}

Equating and dividing by 2^{\,n-4}:

2\binom{n}{3} = \binom{n}{4} \implies \frac{2\,n(n-1)(n-2)}{6} = \frac{n(n-1)(n-2)(n-3)}{24}

Cancelling the non-zero factor n(n-1)(n-2): \;\dfrac{2}{6} = \dfrac{n-3}{24}, so n - 3 = 8 and

n = 11

A1[5]

Final answer

n=11

Question 21

[6 marks]

In the expansion of \dfrac{(ax + 1)^{9}}{6x^{2}}, where a \neq 0, the term in x^{3} has coefficient 14a^{4}.

Determine the value of a.

[6]

Show solutionHide solution

Dividing by 6x^{2} reduces every power of x by 2, so the term in x^{3} of the whole expression comes from the term in x^{5} of (ax + 1)^{9}. That term is

\binom{9}{5}(ax)^{5}(1)^{4} = 126a^{5}x^{5}

Dividing by 6x^{2}:

\frac{126a^{5}x^{5}}{6x^{2}} = 21a^{5}x^{3}

Equating coefficients:

21a^{5} = 14a^{4}

Dividing by a^{4}, which is valid since a \neq 0:

21a = 14 \implies a = \frac{2}{3}

a = \dfrac{2}{3}

[6]

Final answer

a=\frac23

Question 22

[7 marks]

In the binomial expansion

\begin{gathered}(x + 2)^{8} = x^{8} + ax^{7} + bx^{6}\\ {}+ 448x^{5} + dx^{4} + \cdots + 256,\\ a, b, d \in \mathbb{Z}^{+},\end{gathered}

(a) Show that d = 1120.

[2]

The fifth term of the expansion is the mean of the fourth term and the sixth term.

(b) Given that x \neq 0, determine the possible values of x.

[5]

Show solutionHide solution

(a) The term in x^{4} takes four factors of 2:

dx^{4} = \binom{8}{4}x^{4}\,2^{4} \implies d = 70 \times 16 = 1120

A1[2]

(b) The fourth, fifth and sixth terms are

448x^{5}, \qquad 1120x^{4}, \qquad \binom{8}{5}\,2^{5}x^{3} = 56 \times 32\,x^{3} = 1792x^{3}

The fifth term is their mean:

1120x^{4} = \frac{448x^{5} + 1792x^{3}}{2} \implies 448x^{5} - 2240x^{4} + 1792x^{3} = 0

Dividing by 448x^{3} (as x \neq 0): \;x^{2} - 5x + 4 = 0, so (x-1)(x-4) = 0 and

x = 1 \quad \text{or} \quad x = 4

A1[5]

Final answer

(a) shown
(b) x=1 or x=4

Question 23

[7 marks]

In the expansion of (1 + hx)^{9}, where h > 0, the coefficients of x, x^{2} and x^{4} are, in that order, three consecutive terms of a geometric sequence.

Determine h.

[7]

Show solutionHide solution

Coefficients: \binom{9}{1}h = 9h, \ \binom{9}{2}h^2 = 36h^2, \ \binom{9}{4}h^4 = 126h^4

M1A1A1

Geometric: (middle)^2 = (first)(third)

\left(36h^2\right)^2 = (9h)(126h^4)

1296h^4 = 1134h^5

Since h > 0: h = \dfrac{1296}{1134} = \dfrac{8}{7}

A1[7]

Final answer

h=\frac87

Question 24

[7 marks]

Asha puts 3600 rupees into a savings account that pays 5\% compound interest per year. After three full years the balance is 3600(1 + k)^{3}, where k \in \mathbb{Q}.

(a) State the value of k.

[1]

(b) Write (1 + x)^{3} as a polynomial in x.

[2]

(c) Use your answer to (b) to calculate the balance after three years, to the nearest rupee.

[4]

Show solutionHide solution

(a) k = \dfrac{5}{100} = 0.05

[1]

(b) (1 + x)^{3} = 1 + 3x + 3x^{2} + x^{3}

[2]

(c) Replacing x by 0.05:

(1.05)^{3} = 1 + 3(0.05) + 3(0.05)^{2} + (0.05)^{3}

= 1 + 0.15 + 0.0075 + 0.000125 = 1.157625

3600 \times 1.157625 = 4167.45 \approx \mathbf{4167 \text{ rupees}}

A1[4]

Final answer

(a) k=0.05
(b) 1+3x+3x^2+x^3
(c) 4167 rupees

Question 25

[6 marks]

When \dfrac{1}{\sqrt[3]{q - x^{2}}} is expanded, where q \in \mathbb{Q}^{+}, the coefficient of x^{4} is \dfrac{1}{576}.

Determine the value of q.

[6]

Show solutionHide solution

Taking out the factor q so that the binomial series applies:

\frac{1}{\sqrt[3]{q - x^{2}}} = \left(q - x^{2}\right)^{-\frac{1}{3}} = q^{-\frac{1}{3}}\left(1 - \frac{x^{2}}{q}\right)^{-\frac{1}{3}}

In the expansion of (1 + u)^{-\frac{1}{3}} the term in u^{2} is

\frac{\left(-\frac{1}{3}\right)\left(-\frac{1}{3} - 1\right)}{2!}\,u^{2} = \frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2}\,u^{2} = \frac{2}{9}u^{2}

Putting u = -\dfrac{x^{2}}{q} gives u^{2} = \dfrac{x^{4}}{q^{2}}, so the coefficient of x^{4} is

q^{-\frac{1}{3}} \times \frac{2}{9} \times \frac{1}{q^{2}} = \frac{2}{9}q^{-\frac{7}{3}}

Setting this equal to \dfrac{1}{576}:

\frac{2}{9}q^{-\frac{7}{3}} = \frac{1}{576} \implies q^{\frac{7}{3}} = \frac{2 \times 576}{9} = 128

q = 128^{\frac{3}{7}} = \left(2^{7}\right)^{\frac{3}{7}} = 2^{3} = 8

q = 8

[6]

Final answer

q=8

Question 26

[7 marks]

Let E(x) = \sqrt[3]{1 + ax} - \dfrac{1}{1 + x}, where a is a non-zero rational number. Up to and including the term in x^{2}, the binomial expansion of E(x) is bx - bx^{2} for some rational b.

(a) Find a and b.

[6]

(b) For which values of x is this expansion valid?

[1]

Show solutionHide solution

(a) Expanding each part:

(1 + ax)^{1/3} = 1 + \frac{1}{3}(ax) + \frac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2!}(ax)^{2} + \cdots = 1 + \frac{a}{3}x - \frac{a^{2}}{9}x^{2} + \cdots

(1 + x)^{-1} = 1 - x + x^{2} - \cdots

Subtracting:

\sqrt[3]{1 + ax} - \frac{1}{1 + x} = \left(\frac{a}{3} + 1\right)x - \left(\frac{a^{2}}{9} + 1\right)x^{2} + \cdots

Matching with bx - bx^{2} gives b = \dfrac{a}{3} + 1 and b = \dfrac{a^{2}}{9} + 1. Hence

\frac{a}{3} = \frac{a^{2}}{9} \implies a(a - 3) = 0

Since a \neq 0: \;a = 3 and b = 2.

A1[6]

(b) The first expansion needs |3x| < 1 and the second needs |x| < 1, so the expansion is valid for

|x| < \frac{1}{3}

A1[1]

Final answer

(a) a=3, b=2
(b) |x|<\frac13

Question 27

[17 marks]

(a) Find the first four terms, in ascending powers of x, of the binomial expansion of \sqrt[4]{1 + 4x}.

[4]

Now consider \dfrac{1 + px}{1 + qx}, where p, q \in \mathbb{Q}.

(b) Expand \dfrac{1 + px}{1 + qx} in ascending powers of x up to and including x^{2}.

[3]

For particular values of p and q, the series in (a) and (b) agree for their first three terms.

(c) Show that q = \dfrac{3}{2}, and find p.

[4]

With these values of p and q, the fraction \dfrac{1 + px}{1 + qx} approximates \sqrt[4]{1 + 4x} when |x| < \dfrac{1}{4}.

(d) (i) Choose a suitable x to obtain an approximation to \sqrt[4]{1.08} as a fraction.

(ii) The same method with x = 0.2 approximates \sqrt[4]{1.8} by \dfrac{15}{13}. Explain why this is less accurate than the approximation in (d)(i).

[6]

Show solutionHide solution

(a) Using the binomial expansion with n = \dfrac{1}{4} and u = 4x:

(1 + 4x)^{1/4} = 1 + \frac{1}{4}(4x) + \frac{\frac{1}{4}\left(-\frac{3}{4}\right)}{2!}(4x)^{2} + \frac{\frac{1}{4}\left(-\frac{3}{4}\right)\left(-\frac{7}{4}\right)}{3!}(4x)^{3} + \cdots

= 1 + x - \frac{3}{32}\left(16x^{2}\right) + \frac{7}{128}\left(64x^{3}\right) + \cdots

= 1 + x - \frac{3}{2}x^{2} + \frac{7}{2}x^{3} + \cdots

A1[4]

(b)

\frac{1 + px}{1 + qx} = (1 + px)(1 + qx)^{-1} = (1 + px)\left(1 - qx + q^{2}x^{2} - \cdots\right)

M1A1

= 1 + (p - q)x + \left(q^{2} - pq\right)x^{2} + \cdots

A1[3]

(c) Comparing the coefficients of x and x^{2} with part (a):

p - q = 1, \qquad q^{2} - pq = -\frac{3}{2}

The second equation factorises as q(q - p) = -\dfrac{3}{2}, and q - p = -1, so

-q = -\frac{3}{2} \implies q = \frac{3}{2}

and then p = q + 1 = \dfrac{5}{2}.

[4]

(d)(i) Taking 1 + 4x = 1.08, that is x = 0.02:

\sqrt[4]{1.08} \approx \frac{1 + \frac{5}{2}(0.02)}{1 + \frac{3}{2}(0.02)} = \frac{1.05}{1.03} = \frac{105}{103}

A1A1

(ii) The series match (1 + 4x)^{1/4} only as far as the term in x^{2}, so the error comes from the terms in x^{3} and beyond.

Those terms grow with |x|. Since x = 0.2 is much further from 0 than x = 0.02 (and is close to the limit \tfrac{1}{4} of validity), the neglected terms are much larger, so \dfrac{15}{13} is the less accurate approximation.

R1R1

(Numerically, \dfrac{105}{103} \approx 1.01942 against \sqrt[4]{1.08} \approx 1.01943, whereas \dfrac{15}{13} \approx 1.15385 against \sqrt[4]{1.8} \approx 1.15829.) [6 for part (d)]

Final answer

(a) 1+x-\frac32x^2+\frac72x^3
(b) 1+(p-q)x+(q^2-pq)x^2
(c) q=\frac32 (shown), p=\frac52
(d)(i) \frac{105}{103}
(ii) x=0.2 is far from 0 and close to the limit \frac14, so the omitted x^3 terms are much larger

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Odd numbers in Pascal's triangle

G1 Worked example

[10 marks]

For polynomials P(x) and Q(x) with integer coefficients, P(x)\equiv Q(x) means that every coefficient of P(x)-Q(x) is even. You may use the fact that if P_1\equiv Q_1 and P_2\equiv Q_2, then P_1P_2\equiv Q_1Q_2.

(a) Show that (1+x)^2\equiv1+x^2.

[1]

(b) Prove by induction that (1+x)^{2^k}\equiv1+x^{2^k} for all k\in\mathbb{Z}^+. Hence state how many entries of row 2^k of Pascal's triangle are odd.

[4]

(c) By writing (1+x)^{12}=(1+x)^8(1+x)^4, find which entries of row 12 are odd.

[3]

(d) Find the number of odd entries in row 13.

[2]

Show solutionHide solution

(a) (1+x)^2-(1+x^2)=2x, and its one coefficient, 2, is even

A1AG

(b) true for k=1 by (a). Assume (1+x)^{2^m}\equiv1+x^{2^m} for some m\in\mathbb{Z}^+

(1+x)^{2^{m+1}}=\left((1+x)^{2^m}\right)^2\equiv\left(1+x^{2^m}\right)^2, using the given fact with P_1=P_2

\left(1+x^{2^m}\right)^2=1+2x^{2^m}+x^{2^{m+1}}\equiv1+x^{2^{m+1}}

true for k=1, and true for k=m implies true for k=m+1, so true for all k\in\mathbb{Z}^+. The entries of row 2^k are the coefficients of (1+x)^{2^k}, so only the first and last are odd: 2 odd entries

(c) (1+x)^{12}\equiv\left(1+x^8\right)\left(1+x^4\right)

=1+x^4+x^8+x^{12}

the odd entries are \binom{12}{0}=1, \binom{12}{4}=495, \binom{12}{8}=495, \binom{12}{12}=1; the other nine are even

(d) 13=8+4+1: (1+x)^{13}\equiv(1+x^8)(1+x^4)(1+x)=1+x+x^4+x^5+x^8+x^9+x^{12}+x^{13}

8 odd entries

Final answer

(b) 2
(c) \binom{12}{0},\binom{12}{4},\binom{12}{8},\binom{12}{12}
(d) 8

G1 Investigation

[13 marks]

Let N(n) be the number of odd entries in row n of Pascal's triangle, that is, among \binom n0,\binom n1,\dots,\binom nn. Use the notation and results of the worked example.

(a) Find N(n) for n=0,1,2,\dots,16.

[2]

(b) Conjecture formulae for N(2^k) and N(2^k-1), for k\in\mathbb{Z}^+.

[2]

(c) The worked example proves the first formula. Use Pascal's rule \binom nr+\binom n{r+1}=\binom{n+1}{r+1}, with n=2^k-1, to prove the second.

[3]

(d) Let 0\le m<2^k. Show that (1+x)^{2^k+m}\equiv(1+x)^m+x^{2^k}(1+x)^m, and explain why no power of x appears in both terms on the right. Deduce that N(2^k+m)=2N(m).

[4]

(e) Hence find N(100). Describe how N(n) can be found from n written in base 2.

[2]

Show solutionHide solution

(a) n=0 to 16: 1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16,2

Note: award A1 only for at least twelve correct values. Only odd or even matters, so each row can be built from the one above with "odd + odd = even".

(b) N(2^k)=2

N(2^k-1)=2^k: every entry of the row is odd

(c) take two neighbouring entries of row 2^k-1. Pascal's rule adds them to an entry of row 2^k that is neither the first nor the last, and every such entry is even by the worked example

so neighbouring entries of row 2^k-1 are both odd or both even

\binom{2^k-1}{0}=1 is odd, so all 2^k entries are odd: N(2^k-1)=2^k

(d) (1+x)^{2^k+m}=(1+x)^{2^k}(1+x)^m\equiv\left(1+x^{2^k}\right)(1+x)^m=(1+x)^m+x^{2^k}(1+x)^m

the first term has powers x^0 to x^m, the second x^{2^k} to x^{2^k+m}, and m<2^k, so the two sets do not overlap

so the coefficients of (1+x)^{2^k+m} have the same parities as \binom m0,\dots,\binom mm (twice over), with every other coefficient even

N(2^k+m)=2N(m)

(e) 100=64+36 and 36=32+4: N(100)=2N(36)=4N(4)=4\times2=8

each 1 in the base-2 form of n doubles the count, starting from N(0)=1: N(n)=2^j, where j is the number of 1s (100=1100100_2 has three, and 2^3=8)

Final answer

(a) 1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16,2
(b) N(2^k)=2, N(2^k-1)=2^k
(e) N(100)=8; N(n)=2^j, j = number of 1s in base 2

G2 The greatest coefficient

G2 Worked example

[9 marks]

Let c_r be the coefficient of x^r in the expansion of (2+3x)^{10}, for r=0,1,\dots,10.

(a) Show that \dfrac{c_{r+1}}{c_r}=\dfrac{3(10-r)}{2(r+1)} for r=0,1,\dots,9.

[2]

(b) Hence find the value of r for which c_r is the greatest coefficient, and state this coefficient.

[3]

(c) In the expansion of (1+kx)^{11}, where k>0, the coefficient of x^4 is greater than every other coefficient. Find the set of possible values of k.

[4]

Show solutionHide solution

(a) c_r=\binom{10}{r}2^{10-r}3^r

\dfrac{c_{r+1}}{c_r}=\dfrac{\binom{10}{r+1}}{\binom{10}{r}}\times\dfrac32=\dfrac{10-r}{r+1}\times\dfrac32=\dfrac{3(10-r)}{2(r+1)}

A1AG

(b) c_{r+1}>c_r\iff30-3r>2r+2\iff r<5.6

so c_0<c_1<\dots<c_6 and c_6>c_7>\dots>c_{10}

r=6; c_6=\binom{10}{6}2^43^6=210\times16\times729=2\,449\,440

(c) neighbour ratio \dfrac{(11-r)k}{r+1}; it falls as r rises, so c_4 is the greatest exactly when c_4>c_3 and c_4>c_5

\dfrac{c_4}{c_3}=\dfrac{8k}{4}=2k>1

\dfrac{c_5}{c_4}=\dfrac{7k}{5}<1

\frac12<k<\frac57

Final answer

(b) r=6, c_6=2\,449\,440
(c) \frac12<k<\frac57

G2 Investigation

[13 marks]

For a,b>0 and n\in\mathbb{Z}^+, let c_r be the coefficient of x^r in the expansion of (a+bx)^n.

(a) For each of (1+2x)^7, (2+x)^9 and (3+5x)^8, find the value of r for which c_r is greatest.

[3]

(b) For each expansion in (a), calculate \dfrac{(n+1)b}{a+b}. Conjecture a rule that gives the value of r for which c_r is greatest.

[2]

(c) Show that c_{r+1}>c_r if and only if r+1<\dfrac{(n+1)b}{a+b}.

[3]

(d) Hence prove your conjecture in the case when \dfrac{(n+1)b}{a+b} is not an integer.

[2]

(e) For the expansion of (1+kx)^{20}, where k>0, find the values of k for which two coefficients are equal and greater than all the others.

[3]

Show solutionHide solution

(a) (1+2x)^7: r=5 (c_5=672)

(2+x)^9: r=3 (c_3=5376)

(3+5x)^8: r=5 (c_5=4\,725\,000)

(b) \dfrac{16}3=5.33\ldots, \dfrac{10}3=3.33\ldots, \dfrac{45}8=5.625

conjecture: the greatest coefficient is c_r where r is the integer part of \dfrac{(n+1)b}{a+b}

(c) \dfrac{c_{r+1}}{c_r}=\dfrac{\binom n{r+1}a^{n-r-1}b^{r+1}}{\binom nra^{n-r}b^r}=\dfrac{(n-r)b}{(r+1)a}

M1A1

c_{r+1}>c_r\iff(n-r)b>(r+1)a\iff(n+1)b>(r+1)(a+b)\iff r+1<\dfrac{(n+1)b}{a+b} (all quantities positive)

A1AG

(d) let R=\dfrac{(n+1)b}{a+b}, not an integer, and q its integer part, so q<R<q+1. By (c), c_{r+1}>c_r for r+1\le q and c_{r+1}<c_r for r+1\ge q+1 (equality would need r+1=R)

so c_0<\dots<c_q and c_q>c_{q+1}>\dots>c_n: the greatest is c_q

(e) by (c), c_{r+1}=c_r exactly when r+1=R=\dfrac{21k}{1+k}, so R must be a whole number m

0<\dfrac{21k}{1+k}<21, so m\in\{1,2,\dots,20\}, and then c_{m-1}=c_m is greater than every other coefficient

k=\dfrac{m}{21-m} for m=1,2,\dots,20, that is k=\frac1{20},\frac2{19},\frac3{18},\dots,20

Note: for example k=\frac12 (m=7) gives c_6=c_7=605.625 in \left(1+\frac x2\right)^{20}.

Final answer

(a) 5, 3, 5
(b) r= integer part of \frac{(n+1)b}{a+b}
(e) k=\frac{m}{21-m}, m=1,2,\dots,20

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

Photon Guidebook: Mathematics Analysis and Approaches HL for the IB Diploma

Chapter 3 teaches binomial theorem from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book