Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 3: The binomial theorem and proof worked solutions

Worked solutions to the practice questions in Chapter 3 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL (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

[6 marks]
For integers 0\le r\le n, \dbinom{n}{r}=\dfrac{n!}{r!\,(n-r)!}, where n!=n\times(n-1)\times\dots\times 2\times 1 and 0!=1.
(a) Use the formula to show that \dbinom{7}{3}=35.
[2]
(b) Find the value of \dbinom{10}{2}.
[2]
(c) Find the value of \dbinom{12}{10}.
[2]
Show solutionHide solution
(a) \dbinom{7}{3}=\dfrac{7!}{3!\,4!}
M1
=\dfrac{7\times 6\times 5}{3\times 2\times 1}=\dfrac{210}{6}=35
A1AG
(b) \dfrac{10!}{2!\,8!}=\dfrac{10\times 9}{2}
M1
=45
A1
(c) \dfrac{12!}{10!\,2!}=\dfrac{12\times 11}{2}
M1
=66
A1

Final answer

  • (a) 35
  • (b) 45
  • (c) 66

Question 2

[6 marks]
Expand and simplify
(a) (2x+3)^3;
[3]
(b) (3-x)^4.
[3]
Show solutionHide solution
(a) (2x)^3+3(2x)^2(3)+3(2x)(3)^2+3^3
M1
=8x^3+36x^2+54x+27
A2
Note: award A1 for three correct terms.
(b) 3^4+4(3)^3(-x)+6(3)^2(-x)^2+4(3)(-x)^3+(-x)^4
M1
=81-108x+54x^2-12x^3+x^4
A2
Note: award A1 for four correct terms.

Final answer

  • (a) 8x^3+36x^2+54x+27
  • (b) 81-108x+54x^2-12x^3+x^4

Question 3

[6 marks]
(a) Expand \left(2x-\dfrac{1}{x}\right)^4, simplifying each term.
[3]
(b) Write down the term independent of x in this expansion.
[1]
(c) Hence simplify \left(2x-\dfrac{1}{x}\right)^4+\left(2x+\dfrac{1}{x}\right)^4.
[2]
Show solutionHide solution
(a) (2x)^4+4(2x)^3\left(-\dfrac{1}{x}\right)+6(2x)^2\left(-\dfrac{1}{x}\right)^2+4(2x)\left(-\dfrac{1}{x}\right)^3+\left(-\dfrac{1}{x}\right)^4
M1
=16x^4-32x^2+24-\dfrac{8}{x^2}+\dfrac{1}{x^4}
A2
Note: award A1 for three correct terms.
(b) 24
A1
(c) \left(2x+\dfrac{1}{x}\right)^4=16x^4+32x^2+24+\dfrac{8}{x^2}+\dfrac{1}{x^4} (the terms with odd powers of -\dfrac{1}{x} change sign)
M1
sum =32x^4+48+\dfrac{2}{x^4}
A1

Final answer

  • (a) 16x^4-32x^2+24-8x^{-2}+x^{-4}
  • (b) 24
  • (c) 32x^4+48+2x^{-4}

Question 4

[6 marks]
(a) For (1+5x)^8, write the terms in x^0, x and x^2, in that order.
[3]
(b) Expand (2x-1)^6 as far as x^4, starting with its highest power.
[3]
Show solutionHide solution
(a) 1+\dbinom{8}{1}(5x)+\dbinom{8}{2}(5x)^2
M1
=1+40x+700x^2 A1A1
Note: A1 for 40x, A1 for 700x^2.
(b) (2x)^6+\dbinom{6}{1}(2x)^5(-1)+\dbinom{6}{2}(2x)^4(-1)^2
M1
=64x^6-192x^5+240x^4 A1A1
Note: A1 for -192x^5, A1 for 240x^4.

Final answer

  • (a) 1+40x+700x^2
  • (b) 64x^6-192x^5+240x^4

Question 5

[5 marks]
In the expansion of (3+ax)^4, where a\ne 0, the sum of the coefficient of x and the coefficient of x^2 is zero.
(a) Find the value of a.
[4]
(b) Find the coefficient of x^3.
[1]
Show solutionHide solution
(a) coefficient of x: \dbinom{4}{1}3^3a=108a (A1)
coefficient of x^2: \dbinom{4}{2}3^2a^2=54a^2 (A1)
108a+54a^2=0 \Rightarrow 54a(a+2)=0
M1
a\ne 0, so a=-2
A1
(b) \dbinom{4}{3}(3)(-2)^3=-96
A1

Final answer

  • (a) a=-2
  • (b) -96

Question 6

[4 marks]
(a) Show that (x-6)^2+5\equiv x^2-12x+41.
[2]
(b) By starting with the right-hand side, show that 5x^2+30x+38\equiv 5(x+3)^2-7.
[2]
Show solutionHide solution
(a) LHS =x^2-12x+36+5
M1
=x^2-12x+41= RHS
A1
(b) RHS =5(x^2+6x+9)-7
M1
=5x^2+30x+45-7=5x^2+30x+38= LHS
A1
Note: award M1 for a correct expansion of the square; the final line must reach the other side exactly.

Final answer

  • (a) x^2-12x+41
  • (b) 5x^2+30x+38

Question 7

[4 marks]
(a) Let m be an integer. Write down an expression in terms of m for an even integer and for an odd integer.
[1]
(b) Prove that the product of any two odd integers is odd.
[3]
Show solutionHide solution
(a) even: 2m; odd: 2m+1
A1
(b) let the integers be 2a+1 and 2b+1, where a,\ b are integers
M1
Note: using the same letter for both ((2a+1)^2) only proves the result for a square: award M0.
(2a+1)(2b+1)=4ab+2a+2b+1
A1
=2(2ab+a+b)+1, and 2ab+a+b is an integer, so the product is odd
R1

Final answer

  • (a) 2m, 2m+1
  • (b) 2(2ab+a+b)+1, odd

Question 8

[6 marks]
A two-digit number has tens digit a and units digit b, where a and b are whole numbers from 1 to 9. Reversing the digits gives another two-digit number.
(a) Show that 73-37 is a multiple of 9.
[1]
(b) Explain why the number can be written as 10a+b.
[1]
(c) Prove that the difference between the number and its reversal is always a multiple of 9.
[2]
(d) Prove that the sum of the number and its reversal is always a multiple of 11.
[2]
Show solutionHide solution
(a) 73-37=36=9\times 4
A1
(b) the tens digit is worth 10 times its value and the units digit its own value, so the number is 10a+b
R1
(c) (10a+b)-(10b+a)=9a-9b
M1
=9(a-b), and a-b is an integer
A1
(d) (10a+b)+(10b+a)=11a+11b
M1
=11(a+b), and a+b is an integer
A1

Final answer

  • (a) 36=9(4)
  • (c) 9(a-b)
  • (d) 11(a+b)

Tutorial

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

Question 9

[6 marks]
(a) Expand (3+\sqrt2)^4, writing the result as p+q\sqrt2, for integer p and q.
[3]
(b) Hence write down the value of (3-\sqrt2)^4 in the form p-q\sqrt2.
[1]
(c) Hence find the value of (3+\sqrt2)^4(3-\sqrt2)^4.
[2]
Show solutionHide solution
(a) (3+\sqrt2)^4=3^4+4(3)^3\sqrt2+6(3)^2(\sqrt2)^2+4(3)(\sqrt2)^3+(\sqrt2)^4
[1]
=81+108\sqrt2+6(9)(2)+12(2\sqrt2)+4
[1]
=193+132\sqrt2.
[1]
(b) Changing the sign of \sqrt2 changes the sign of the odd powers: (3-\sqrt2)^4=193-132\sqrt2.
[1]
(c) (193+132\sqrt2)(193-132\sqrt2)=193^2-2(132)^2
[1]
=37\,249-34\,848=2401.
[1]
(Check: (3+\sqrt2)(3-\sqrt2)=7 and 7^4=2401.)

Final answer

  • (a) 193+132\sqrt2
  • (b) 193-132\sqrt2
  • (c) 2401

Question 10

[6 marks]
(a) Obtain the constant term when you expand \left(\sqrt{x}+\dfrac2x\right)^9.
[4]
(b) Which number multiplies x^{-3} in your expansion?
[2]
Show solutionHide solution
(a) T_{r+1}=\dbinom9r\left(x^{1/2}\right)^{9-r}\left(2x^{-1}\right)^r=\dbinom9r2^rx^{\frac{9-r}{2}-r}=\dbinom9r2^rx^{\frac{9-3r}{2}}.
[2]
Independent of x: \dfrac{9-3r}{2}=0\Rightarrow r=3.
[1]
Term =\dbinom93\times2^3=84\times8=672.
[1]
(b) \dfrac{9-3r}{2}=-3\Rightarrow9-3r=-6\Rightarrow r=5.
[1]
Coefficient =\dbinom95\times2^5=126\times32=4032.
[1]

Final answer

  • (a) 672
  • (b) 4032

Question 11

[6 marks]
In the expansion of (1-3x)^n, where n\in\mathbb{Z}^+, the coefficient of x^2 is 252.
(a) Find the value of n.
[4]
(b) Find the sum of all the coefficients in the expansion.
[2]
Show solutionHide solution
(a) Coefficient of x^2: \dbinom n2\times(-3)^2=\dfrac{9n(n-1)}{2}.
[2]
\dfrac{9n(n-1)}{2}=252\Rightarrow n^2-n-56=0.
[1]
(n-8)(n+7)=0\Rightarrow n=8 (reject -7).
[1]
(b) The sum of the coefficients is the value at x=1: (1-3)^8
[1]
=(-2)^8=256.
[1]

Final answer

  • (a) n=8
  • (b) 256

Question 12

[5 marks]
In the expansion of (3+2x)^n, where n\in\mathbb{Z}^+ and n\ge3, write c_2 and c_3 for the coefficients of x^2 and x^3. Given c_3=2c_2, determine n.
Show solutionHide solution
Coefficient of x^2: \dbinom n2\,3^{n-2}\,2^2.
[1]
Coefficient of x^3: \dbinom n3\,3^{n-3}\,2^3.
[1]
Equate (coefficient of x^3 =2\times coefficient of x^2): \dbinom n3\,3^{n-3}\cdot8=2\dbinom n2\,3^{n-2}\cdot4.
[1]
Use \dbinom n3=\dbinom n2\times\dfrac{n-2}{3} and cancel \dbinom n2\,3^{n-3}: \dfrac{n-2}{3}\times8=2\times3\times4.
[1]
8(n-2)=72\Rightarrow n=11.
[1]
(Check: for n=11 the ratio of the two coefficients is \dfrac{165\times8}{55\times3\times4}=2.)

Final answer

n=11

Question 13

[6 marks]
Use your GDC for this question.
For (1+6x)^n, the number multiplying x^3, where n\in\mathbb{Z}^+, is 18\,144.
(a) Determine n.
[4]
(b) Using your index, calculate the coefficient multiplying x^5.
[2]
Show solutionHide solution
(a) Coefficient of x^3: \dbinom n3\times6^3=216\dbinom n3.
[2]
216\dbinom n3=18\,144\Rightarrow\dbinom n3=84.
[1]
Enter y=\binom{x}{3} and scroll the table (or solve \frac{x(x-1)(x-2)}{6}=84 on the GDC): n=9.
[1]
(b) \dbinom{9}{5}\times6^5=126\times7776
[1]
=979\,776.
[1]

Final answer

  • (a) n=9
  • (b) 979\,776

Question 14

[4 marks]
(a) Prove that (2n+7)^2 - (2n+1)^2 is a multiple of 24 for every integer n.
[3]
(b) Find the smallest positive value that this expression takes as n ranges over the integers.
[1]
Show solutionHide solution
(a) Difference of two squares: \bigl[(2n+7)-(2n+1)\bigr]\bigl[(2n+7)+(2n+1)\bigr].
[1]
= 6(4n+8) = 24(n+2).
[1]
Since n+2 is an integer, the expression is a multiple of 24.
[1]
(b) 24(n+2) is smallest and positive when n+2 = 1, i.e. n=-1, giving 24.
[1]

Final answer

  • (a) 24(n+2)
  • (b) 24, at n=-1

Question 15

[4 marks]
Prove that n^2 + 5n is even for every integer n, by considering the cases n even and n odd.
Show solutionHide solution
Every integer is even or odd, so these two cases are exhaustive.
[1]
Case 1: n = 2k. Then n^2+5n = 4k^2 + 10k = 2(2k^2+5k), which is even.
[1]
Case 2: n = 2k+1. Then n^2+5n = 4k^2+4k+1+10k+5 = 4k^2+14k+6 = 2(2k^2+7k+3), which is even.
[1]
In both cases n^2+5n is even, and the cases cover all integers.
[1]

Final answer

2(2k^2+5k) if n=2k; 2(2k^2+7k+3) if n=2k+1

Question 16

[5 marks]
A rectangular mosaic has 2m+1 rows and 2m+3 columns of unit tiles, where m is a non-negative integer. Its area is A square units.
(a) Show that A=4(m+1)^2-1.
[3]
(b) Hence prove that A+1 is a square number divisible by 4.
[2]
Show solutionHide solution
(a) A=(2m+1)(2m+3)
M1
=4m^2+8m+3=4(m+1)^2-1.
A1A1
(b) A+1=[2(m+1)]^2, the square of an integer.
A1
Also A+1=4(m+1)^2, where (m+1)^2 is an integer, so it is divisible by 4.
R1

Final answer

  • (a) 4(m+1)^2-1
  • (b) A+1=[2(m+1)]^2=4(m+1)^2

Mastery

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

Question 17

[5 marks]
Let a\in\mathbb{R}, a\neq0.
(a) Expand (2+ax)^5 only far enough to obtain its x^3 coefficient in terms of a.
[2]
(b) For (1+3ax)^6, obtain the coefficient multiplying x^2, in terms of a.
[1]
(c) The two coefficients are equal. Find the value of a.
[2]
Show solutionHide solution
(a) \dbinom53\times2^2\times a^3
[1]
=40a^3.
[1]
(b) \dbinom62\times(3a)^2=15\times9a^2=135a^2.
[1]
(c) 40a^3=135a^2\Rightarrow5a^2(8a-27)=0.
[1]
Since a\neq0, a=\dfrac{27}{8}.
[1]

Final answer

  • (a) 40a^3
  • (b) 135a^2
  • (c) a=27/8

Question 18

[6 marks]
In the expansion of \left(kx^2+\dfrac1x\right)^6, where k\in\mathbb{R} and k\neq0, write c_3 for the coefficient of x^3 and c_0 for the constant term. They satisfy c_3=4c_0.
Find the value of k.
Show solutionHide solution
T_{r+1}=\dbinom6r(kx^2)^{6-r}(x^{-1})^r=\dbinom6rk^{6-r}x^{12-3r}.
[2]
Term in x^3: 12-3r=3\Rightarrow r=3, coefficient \dbinom63k^3=20k^3.
[1]
Constant term: 12-3r=0\Rightarrow r=4, value \dbinom64k^2=15k^2.
[1]
20k^3=4\times15k^2.
[1]
20k^2(k-3)=0, and k\neq0, so k=3.
[1]

Final answer

k=3

Question 19

[7 marks]
(a) Show that n^3 + 23n = (n-1)n(n+1) + 24n.
[1]
(b) By considering the cases n = 3k, n = 3k+1 and n = 3k+2, prove that (n-1)n(n+1) is divisible by 3 for every integer n.
[3]
(c) Hence prove that n^3 + 23n is divisible by 6 for every integer n. You may use the fact that the product of two consecutive integers is even.
[3]
Show solutionHide solution
(a) (n-1)n(n+1) = n\left(n^2-1\right) = n^3 - n, so (n-1)n(n+1) + 24n = n^3 - n + 24n = n^3 + 23n.
[1]
(b) If n=3k, the factor n is a multiple of 3.
[1]
If n=3k+1 then n-1=3k is a multiple of 3; if n=3k+2 then n+1=3(k+1) is a multiple of 3.
[1]
These cases cover every integer and each makes one factor a multiple of 3, so (n-1)n(n+1) is divisible by 3.
[1]
(c) The product (n-1)n is a product of two consecutive integers, hence even, so (n-1)n(n+1) is even; with (b) it is a multiple of both 2 and 3, hence of 6.
[2]
24n = 6(4n) is a multiple of 6, so n^3+23n = (n-1)n(n+1) + 24n is divisible by 6.
[1]

Final answer

  • (a) n^3+23n
  • (b) one factor is divisible by 3 in each case
  • (c) the consecutive product and 24n are divisible by 6

Question 20

[6 marks]
Consider the expansion of (2 + x)(1 + ax)^{5}, where a \in \mathbb{R}. The coefficient of x^{2} is 70. Determine the possible values of a.
Show solutionHide solution
Expanding the bracket to the term in x^{2}:
M1
(1 + ax)^{5} = 1 + 5ax + 10a^{2}x^{2} + \cdots
A1
The x^{2} term of the product arises from 2 \times 10a^{2}x^{2} and x \times 5ax:
M1
\left(20a^{2} + 5a\right)x^{2}
A1
Equating to 70:
M1
20a^{2} + 5a - 70 = 0 \implies 4a^{2} + a - 14 = 0 \implies (4a - 7)(a + 2) = 0
a = \frac{7}{4} \quad \text{or} \quad a = -2
A1[6]

Final answer

a=\frac74 or a=-2

Question 21

[7 marks]
Let a,b,c be the positive integer coefficients multiplying x,x^2,x^3, respectively, in (1+3x)^7.
(a) Show that c=945.
[2]
(b) For x\ne0, the middle entry of ax,bx^2,cx^3 is the arithmetic mean of the other two. Determine all possible values of x.
[5]
Show solutionHide solution
(a) The term in x^{3} is
\binom{7}{3}(3x)^{3} = 35 \times 27x^{3} = 945x^{3}
M1
so c = 945.
A1[2]
(b) Similarly a = 7 \times 3 = 21 and b = \binom{7}{2} \times 9 = 189.
M1
The mean condition gives
M1
189x^{2} = \frac{21x + 945x^{3}}{2} \implies 945x^{3} - 378x^{2} + 21x = 0
A1
Dividing by 21x:
M1
45x^{2} - 18x + 1 = 0 \implies (3x - 1)(15x - 1) = 0
x = \frac{1}{3} \quad \text{or} \quad x = \frac{1}{15}
A1[5]

Final answer

  • (a) 945
  • (b) x=1/3 or x=1/15

Question 22

[5 marks]
In the expansion of \left(\sqrt{x} + \dfrac{k}{x}\right)^{9}, where k \in \mathbb{Z}, the constant term is -2268. Determine the value of k.
[5]
Show solutionHide solution
The general term is
M1
\binom{9}{r}\left(x^{1/2}\right)^{9-r}\left(\frac{k}{x}\right)^{r} = \binom{9}{r}k^{r}\,x^{\frac{9-r}{2}-r} = \binom{9}{r}k^{r}\,x^{\frac{9-3r}{2}}
A1
A constant term needs \dfrac{9 - 3r}{2} = 0, so r = 3:
A1
\binom{9}{3}k^{3} = 84k^{3}
Equating:
M1
84k^{3} = -2268 \implies k^{3} = -27
k = -3
A1[5]

Final answer

k=-3

Question 23

[6 marks]
Three labels carry the numbers 100a+10b+c, 100b+10c+a and 100c+10a+b, where each of a,b,c is a digit from 1 to 9.
(a) Show that their total is 111(a+b+c).
[2]
(b) Prove that this total is divisible by 37.
[2]
(c) The total is 1332, with a=3 and b=4. Find c.
[2]
Show solutionHide solution
(a) Total =(100+10+1)a+(10+1+100)b+(1+100+10)c
M1
=111(a+b+c).
A1
(b) 111=37\times3, so total =37[3(a+b+c)].
M1
3(a+b+c) is an integer, hence the total is divisible by 37.
R1
(c) 111(3+4+c)=1332, so 3+4+c=12.
M1
c=5.
A1

Final answer

  • (a) 111(a+b+c)
  • (b) 37[3(a+b+c)]
  • (c) c=5

Question 24

[10 marks]
Let F(x)=(2+x)^4+(2-x)^4.
(a) Expand (2+x)^4.
[3]
(b) Show that F(x)=32+48x^2+2x^4.
[2]
(c) Obtain F(3) exactly.
[2]
(d) For odd integer x, prove that F(x) is divisible by 2 but not by 4.
[3]
Show solutionHide solution
(a) Use row 1,4,6,4,1 of Pascal's triangle.
M1
(2+x)^4=16+32x+24x^2+8x^3+x^4.
A1A1
(b) Changing x to -x reverses the odd-power signs, so they cancel when the two expansions are added.
R1
F(x)=32+48x^2+2x^4.
A1
(c) F(3)=32+48(9)+2(81)
M1
=626.
A1
(d) F(x)=2(x^4+24x^2+16).
M1
For odd x, x^4 is odd while 24x^2+16 is even; hence the bracket is odd.
A1
Twice an odd integer is divisible by 2 and has remainder 2 when divided by 4.
R1

Final answer

  • (a) 16+32x+24x^2+8x^3+x^4
  • (b) 32+48x^2+2x^4
  • (c) 626
  • (d) twice an odd integer

Going further

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

G1 The greatest coefficient

G1 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 r for which c_r is greatest, 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
M1
\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
M1
so c_0<c_1<\dots<c_6 and c_6>c_7>\dots>c_{10}
R1
r=6; c_6=\binom{10}{6}2^43^6=210\times16\times729=2\,449\,440
A1
(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
M1
\dfrac{c_4}{c_3}=\dfrac{8k}{4}=2k>1
A1
\dfrac{c_5}{c_4}=\dfrac{7k}{5}<1
A1
\frac12<k<\frac57
A1

Final answer

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

G1 Investigation

[13 marks]
Let c_r denote the coefficient multiplying x^r in (a+bx)^n, where a,b>0 and n is a positive integer. Write R=(n+1)b/(a+b).
(a) Find the index r of the greatest coefficient in (1+2x)^7, (2+x)^9 and (3+5x)^8.
[3]
(b) Calculate R for these three cases and conjecture an index rule.
[2]
(c) Show that c_{r+1}>c_r exactly when r+1<R.
[3]
(d) Prove your rule for non-integer R.
[2]
(e) Determine all k>0 for which (1+kx)^{20} has two equal greatest coefficients.
[3]
Show solutionHide solution
(a) (1+2x)^7: r=5 (c_5=672)
A1
(2+x)^9: r=3 (c_3=5376)
A1
(3+5x)^8: r=5 (c_5=4\,725\,000)
A1
(b) \dfrac{16}3=5.33\ldots, \dfrac{10}3=3.33\ldots, \dfrac{45}8=5.625
A1
conjecture: the greatest coefficient is c_r where r is the integer part of \dfrac{(n+1)b}{a+b}
A1
(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)
M1
so c_0<\dots<c_q and c_q>c_{q+1}>\dots>c_n: the greatest is c_q
R1
(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
M1
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
R1
k=\dfrac{m}{21-m} for m=1,2,\dots,20, that is k=\frac1{20},\frac2{19},\frac3{18},\dots,20
A1
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

G2 Cauchy–Schwarz from a discriminant

G2 Worked example

[5 marks]
Use the Cauchy–Schwarz inequality, \left(a_1b_1+a_2b_2\right)^2\le\left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right), with equality exactly when (a_1,a_2) is a multiple of (b_1,b_2).
(a) Real numbers x and y satisfy x+y=5. By taking a_1=x\sqrt2, a_2=y\sqrt3, b_1=\dfrac1{\sqrt2} and b_2=\dfrac1{\sqrt3}, show that 2x^2+3y^2\ge30.
[3]
(b) Find the values of x and y for which 2x^2+3y^2=30.
[2]
Show solutionHide solution
(a) a_1b_1+a_2b_2=x+y=5
M1
a_1^2+a_2^2=2x^2+3y^2 and b_1^2+b_2^2=\frac12+\frac13=\frac56
A1
25\le\left(2x^2+3y^2\right)\times\frac56, so 2x^2+3y^2\ge30
A1AG
(b) equality when \left(x\sqrt2,y\sqrt3\right)=\lambda\left(\frac1{\sqrt2},\frac1{\sqrt3}\right), that is 2x=3y=\lambda
M1
\frac\lambda2+\frac\lambda3=5 gives \lambda=6: x=3, y=2
A1

Final answer

  • (a) 2x^2+3y^2\ge30
  • (b) x=3, y=2

G2 Investigation

[5 marks]
This investigation proves the Cauchy–Schwarz inequality for two pairs by expanding a square.
(a) For a=(1,3), b=(2,5), and then for a=(2,-1), b=(4,-2), compare \left(a_1b_1+a_2b_2\right)^2 with \left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right).
[2]
(b) Show that \left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)-\left(a_1b_1+a_2b_2\right)^2=\left(a_1b_2-a_2b_1\right)^2. Hence prove the inequality for two pairs, and state when equality holds.
[3]
Show solutionHide solution
(a) (2+15)^2=289 and (1+9)(4+25)=290: just below
A1
(8+2)^2=100 and 5\times20=100: equal, and here b=2a
A1
(b) the product is a_1^2b_1^2+a_1^2b_2^2+a_2^2b_1^2+a_2^2b_2^2 and the square is a_1^2b_1^2+2a_1a_2b_1b_2+a_2^2b_2^2
M1
difference =a_1^2b_2^2-2a_1a_2b_1b_2+a_2^2b_1^2=\left(a_1b_2-a_2b_1\right)^2
A1
a square is never negative, so the inequality holds; equality exactly when a_1b_2=a_2b_1, that is when one pair is a multiple of the other
R1

Final answer

  • (a) 289<290, 100=100
  • (b) difference (a_1b_2-a_2b_1)^2\ge0, equality when a_1b_2=a_2b_1

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 12.

JobTI-84 Plus CE keystrokes
Solve an equationMATH, 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

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL

Chapter 3 teaches the binomial theorem and proof 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

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