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Binomial Theorem

Practice exam-style IB Math AA questions for Binomial Theorem, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

Consider the expansion of (2−3x)5(2-3x)^5 in ascending powers of xx.

A

the coefficient of x2x^2.

[2]
B

the first three terms of the expansion in ascending powers of xx.

[3]
Question 2
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

In the expansion of (1+kx)6(1+kx)^6, where kk is a positive constant, the coefficient of x2x^2 is 6060.

A

k=2k=2.

[3]
B

the coefficient of x3x^3.

[2]
Question 3
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

Consider the expansion of (2−3x)8(2-3x)^8.

A

the number of terms in the expansion.

[1]
B

the coefficient of x2x^2.

[3]
Question 4
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

A row of Pascal's triangle is numbered nn, starting with row 00. In one row, the second entry is 1111.

A

the value of nn.

[1]
B

Using a GDC or otherwise, all values of rr such that (11r)=330\binom{11}{r}=330.

[2]
C

the coefficient of a7b4a^7b^4 in the expansion of (a+b)11(a+b)^{11}.

[1]

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Question 5
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The coefficient of x3x^3 in the expansion of (1+2x)n(1+2x)^n is 88 times the coefficient of x2x^2, where n∈Z+n\in\mathbb{Z}^+ and n≥3n\geq 3.

A

expressions, in terms of nn, for the coefficients of x2x^2 and x3x^3.

[3]
B

nn.

[3]
Question 6
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Consider the expansion of

(x+2x2)9\left(x+\frac{2}{x^2}\right)^9

where x≠0x\neq 0.

A

the exponent of xx in the general term corresponding to rr.

[2]
B

the constant term in the expansion.

[4]
Question 7
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The expansion of (1+x)p(1+x)^p in ascending powers of xx begins

1+px+10x2+⋯1+px+10x^2+\cdots

where p∈Z+p\in\mathbb{Z}^+ and p>1p>1.

A

the possible values of pp.

[3]
B

the coefficient of x3x^3 in the expansion.

[2]
Question 8
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The expansion of (1+kx)6(1+kx)^6 is considered, where kk is a real constant.

A

, in terms of kk, the coefficient of x2x^2.

[2]
B

Given that the coefficient of x2x^2 is 240240, the possible values of kk.

[2]
C

For k=4k=4, the coefficient of x5x^5.

[1]

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Question 9
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Consider the expansion of

(x2+3x)9,x≠0\left(x^2+\frac{3}{x}\right)^9, \quad x\ne 0
A

the constant term occurs when r=6r=6.

[2]
B

the constant term.

[3]
Question 10
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the coefficients in the expansion of (1+x)15(1+x)^{15}.

A

Using a GDC table or otherwise, all values of rr for which (15r)>3000\binom{15}{r}>3000.

[2]
B

how many terms in the expansion have coefficients greater than 30003000.

[2]
Question 11
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The first three terms in the expansion of (1+ax)8(1+ax)^8 in ascending powers of xx are

1+24x+bx2+⋯1+24x+bx^2+\cdots

where aa and bb are constants.

A

aa.

[2]
B

bb.

[2]
C

Using your value of aa, the coefficient of x2x^2 in (1+ax)8(1−x)2(1+ax)^8(1-x)^2.

[3]
Question 12
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

In the expansion of (3x−2)7(3x-2)^7, the terms are written in descending powers of xx.

A

the fourth term.

[3]
B

the ratio of the coefficient of x4x^4 to the coefficient of x5x^5.

[3]

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Question 13
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

In the expansion of (1+mx)7(1−2x)(1+mx)^7(1-2x), the coefficient of xx is 55.

A

m=1m=1.

[3]
B

the coefficient of x3x^3 in the expansion of (1+mx)7(1−2x)(1+mx)^7(1-2x).

[3]
Question 14
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the expansion of

(2x2−1x)8\left(2x^2-\frac{1}{x}\right)^8

where x≠0x\neq 0.

A

the general term in the expansion corresponding to rr.

[2]
B

the coefficient of x4x^4.

[4]
Question 15
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A student uses the binomial theorem to approximate (1.01)12(1.01)^{12}.

A

the first three terms in the expansion of (1+x)12(1+x)^{12} in ascending powers of xx.

[3]
B

Use these three terms to approximate (1.01)12(1.01)^{12}.

[1]
C

Using your GDC, the percentage error in this approximation.

[2]
Question 16
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Let n∈Nn\in\mathbb{N}. In the expansion of (3+x2)n(3+x^2)^n, the coefficient of x4x^4 is 12151215.

A

Form an equation in nn.

[3]
B

nn.

[2]

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Question 17
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the polynomial

(1+2x)8(1−x)5(1+2x)^8(1-x)^5
A

the coefficient of x3x^3 in the expansion of (1+2x)8(1+2x)^8.

[1]
B

the coefficient of x4x^4 in the polynomial.

[5]
Question 18
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let a>0a>0. In the expansion of

(ax2−3x)9,x≠0\left(ax^2-\frac{3}{x}\right)^9, \quad x\ne 0

the constant term is 489888489888.

A

the constant term is 61236a361236a^3.

[3]
B

the value of aa.

[2]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

In the expansion of (1+ax)n(1+ax)^n, where a>0a>0 and n∈Nn\in\mathbb{N}, the coefficient of xx is 1818 and the coefficient of x2x^2 is 135135.

A

Form two equations involving aa and nn.

[3]
B

the value of nn and the value of aa.

[3]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For n>4n>4, in the expansion of (2+x)n(2+x)^n, the coefficients of x3x^3 and x4x^4 are equal.

A

the value of nn.

[3]
B

the coefficient of x5x^5.

[2]

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Question 21
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let pp and qq be positive constants. The polynomial

P(x)=(1+px)5(1−qx)3P(x)=(1+px)^5(1-qx)^3

has coefficient of xx equal to 11 and coefficient of x2x^2 equal to −23-23.

A
I.

, in terms of pp and qq, the coefficient of xx in P(x)P(x).

[2]
II.

, in terms of pp and qq, the coefficient of x2x^2 in P(x)P(x).

[3]
B

the values of pp and qq.

[4]
C

the coefficient of x3x^3 in P(x)P(x).

[2]
Question 22
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

In the expansion of (a+bx)7(a+bx)^7 in ascending powers of xx, the first three terms are

128+448x+672x2+⋯128+448x+672x^2+\cdots

where a>0a>0 and b>0b>0.

A
I.

the value of aa.

[2]
II.

the value of bb.

[2]
B

that the coefficient of x2x^2 is 672672.

[2]
C

Using your values of aa and bb, the coefficient of x4x^4 in

(a+bx)7(1−x)2(a+bx)^7(1-x)^2
[4]
Question 23
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

For n∈Z+n\in\mathbb{Z}^+, define

En=(n0)+(n2)22+(n4)24+⋯E_n=\binom{n}{0}+\binom{n}{2}2^2+\binom{n}{4}2^4+\cdots

and

On=(n1)2+(n3)23+(n5)25+⋯O_n=\binom{n}{1}2+\binom{n}{3}2^3+\binom{n}{5}2^5+\cdots

where the sums continue over the possible even and odd indices respectively.

A
I.

an expression for En+OnE_n+O_n.

[1]
II.

an expression for En−OnE_n-O_n.

[2]
B

that

En=3n+(−1)n2E_n=\frac{3^n+(-1)^n}{2}
[4]
C

the value of E6E_6.

[2]
Question 24
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let n∈Z+n\in\mathbb{Z}^+ with n≥3n\geq 3. In the expansion of (1+x)n(1+x)^n, the coefficient of x3x^3 is twice the coefficient of x2x^2.

A
I.

expressions for the coefficients of x2x^2 and x3x^3.

[2]
II.

the value of nn.

[3]
B

Using your value of nn, the coefficient of x4x^4 in

(1+x)n(1−x)2(1+x)^n(1-x)^2
[4]

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Question 25
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The coefficients of three consecutive terms in the expansion of (1+x)n(1+x)^n are 1515, 2020 and 1515, in that order, where n∈Z+n\in\mathbb{Z}^+.

A

Let the coefficient 1515 be (nr)\binom{n}{r}. Write two equations involving nn and rr.

[2]
B

nn and rr.

[5]
Question 26
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The first three terms in the expansion of (a+bx)6(a+bx)^6 in ascending powers of xx are

64+192x+240x2+⋯64+192x+240x^2+\cdots

where a>0a>0 and b>0b>0.

A

aa.

[2]
B

bb.

[3]
C

that the coefficient of x2x^2 is 240240.

[2]
Question 27
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let

S=(n0)+(n2)+(n4)+⋯S=\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots

be the sum of the even-indexed binomial coefficients in the expansion of (1+x)n(1+x)^n, where n∈Z+n\in\mathbb{Z}^+.

A

the value of (1+1)n(1+1)^n as a sum of binomial coefficients.

[2]
B

Using (1−1)n(1-1)^n, that S=2n−1S=2^{n-1}.

[4]
Question 28
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A student uses the binomial theorem to approximate powers of numbers close to 11.

A

Consider the expansion of (1+x)10(1+x)^{10} in ascending powers of xx.

I.

the first four non-zero terms of the expansion of (1+x)10(1+x)^{10}.

[3]
II.

Use your answer to part (a)(i) to approximate (1.025)10(1.025)^{10}.

[2]
B

Using your GDC, the percentage error in the approximation from part (a)(ii).

[3]
C

Let A(x)=1+10x+45x2+120x3A(x)=1+10x+45x^2+120x^3. the largest value of xx, where 0≤x≤0.050\leq x\leq 0.05, for which the percentage error in using A(x)A(x) to approximate (1+x)10(1+x)^{10} is less than 0.010%0.010\%.

[2]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Let

P(x)=(2+px)7(1−qx)3P(x)=(2+px)^7(1-qx)^3

where p>0p>0 and q>0q>0. In the expansion of P(x)P(x), the coefficient of xx is 6464 and the coefficient of x2x^2 is −288-288.

A

Use the given information to pp and qq.

I.

7p−6q=17p-6q=1.

[3]
II.

pp and qq.

[3]
B

For these values of pp and qq, the coefficient of x3x^3 in P(x)P(x).

[3]
C

For these values of pp and qq, the coefficient of x4x^4 in P(x)P(x).

[2]
Question 30
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Let

F(x)=(x2+ax)8F(x)=\left(x^2+\frac{a}{x}\right)^8

where x≠0x\ne 0 and a>0a>0.

A

Consider the general term in the expansion of F(x)F(x) corresponding to the index rr, where 0≤r≤80\leq r\leq 8.

I.

the general term.

[2]
II.

all possible exponents of xx in the expansion.

[2]
B

Given that the coefficient of x4x^4 is 56705670, aa.

[3]
C

Using your value of aa, the coefficient of x−2x^{-2} in F(x)F(x).

[2]
Question 31
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Rows of Pascal's triangle are numbered starting from row 00. The entry in position rr of row nn is (nr)\binom{n}{r}, where r=0,1,…,nr=0,1,\ldots,n.

row

entries

00

11

11

1, 11,\,1

22

1, 2, 11,\,2,\,1

33

1, 3, 3, 11,\,3,\,3,\,1

44

1, 4, 6, 4, 11,\,4,\,6,\,4,\,1

55

1, 5, 10, 10, 5, 11,\,5,\,10,\,10,\,5,\,1

66

1, 6, 15, 20, 15, 6, 11,\,6,\,15,\,20,\,15,\,6,\,1

nn

(n0), (n1), (n2), (n3), (n4), (n5), (n6)\binom{n}{0},\,\binom{n}{1},\,\binom{n}{2},\,\binom{n}{3},\,\binom{n}{4},\,\binom{n}{5},\,\binom{n}{6}

A

In a particular row, the entry in position 44 is five times the entry in position 33.

I.

In the row labelled nn in the table, the entry in position 44 is five times the entry in position 33. the row number is 2323.

[3]
II.

Using a GDC table or otherwise, the largest entry in row 2323.

[1]
B

the coefficient of x4x^4 in the expansion of (2+x)23(2+x)^{23}.

[3]
C

Use a counting argument to Pascal's identity

(n+1r)=(nr−1)+(nr)\binom{n+1}{r}=\binom{n}{r-1}+\binom{n}{r}

where 1≤r≤n1\leq r\leq n.

[3]
Question 32
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A grid route consists of 1212 moves. Each move is either east or north. A route with exactly rr north moves has weight 312−r2r3^{12-r}2^r.

A

Use binomial coefficients to count routes.

I.

the number of routes with exactly rr north moves.

[1]
II.

the total weight of all routes with exactly 55 north moves.

[2]
B

the total weight of all possible routes is 5125^{12}.

[2]
C

A route is selected with probability proportional to its weight. the probability that the selected route has exactly 55 north moves.

[3]

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Question 33
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the product

(x+2x2)7(1−x)8,x≠0\left(x+\frac{2}{x^2}\right)^7(1-x)^8, \quad x\ne 0
A

all pairs (r,s)(r,s) that can contribute to the constant term, where rr is the index in the expansion of (x+2x2)7\left(x+\frac{2}{x^2}\right)^7 and ss is the index in the expansion of (1−x)8(1-x)^8.

[2]
B

the constant term in the product.

[4]
Question 34
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Let kk be a positive constant and let

E(x)=(x2+kx)8(1−x)4,x≠0E(x)=\left(x^2+\frac{k}{x}\right)^8(1-x)^4,\quad x\neq 0

The constant term in the expansion of E(x)E(x) is 1075210752.

A
I.

the general term of (x2+kx)8\left(x^2+\frac{k}{x}\right)^8 corresponding to the index rr, and simplify the power of xx.

[3]
II.

the only contribution to the constant term in E(x)E(x) occurs when r=6r=6 and the term x2x^2 is chosen from (1−x)4(1-x)^4.

[2]
B

the value of kk.

[3]
C

the coefficient of xx in E(x)E(x).

[2]
Question 35
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

Let aa and bb be positive constants. In the expansion of

(1+ax)4(1+bx)4(1+ax)^4(1+bx)^4

the coefficient of xx is 1616 and the coefficient of x2x^2 is 104104.

A
I.

how the coefficient of xx gives a+b=4a+b=4.

[2]
II.

Express the coefficient of x2x^2 in terms of a+ba+b and abab.

[3]
B

the value of abab.

[2]
C

the possible values of aa and bb.

[2]
D

the coefficient of x3x^3 in (1+ax)4(1+bx)4(1+ax)^4(1+bx)^4.

[2]
Question 36
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let n∈Z+n\in\mathbb{Z}^+ and k>0k>0. In the expansion of (1+kx)n(1+kx)^n, the coefficient of xx is 1515 and the coefficient of x2x^2 is 9090.

A
I.

Write an equation using the coefficient of xx.

[1]
II.

Write an equation using the coefficient of x2x^2.

[2]
B

the values of nn and kk.

[4]
C

the coefficient of x3x^3 in (1+kx)n(1+kx)^n.

[2]
D

the coefficient of x3x^3 in

(1+kx)n(1−x)4(1+kx)^n(1-x)^4
[2]

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Question 37
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

Let n∈Z+n\in\mathbb{Z}^+, n≥3n\geq 3. In the expansion of (1+x)n(1+x)^n, the coefficient of x3x^3 is twice the coefficient of x2x^2.

A

Work with the coefficients of x2x^2 and x3x^3 in (1+x)n(1+x)^n.

I.

expressions for these two coefficients.

[2]
II.

nn.

[4]
B

Let

Q(x)=(1+x)8(1−2x)4Q(x)=(1+x)^8(1-2x)^4

the coefficient of x5x^5 in Q(x)Q(x).

[4]
C

Using a GDC, the greatest magnitude of any coefficient in the expansion of (1+x)8(1−2x)4(1+x)^8(1-2x)^4.

[2]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let n∈Z+n\in\mathbb{Z}^+ and k>0k>0. In the expansion of (1+kx)n(1+kx)^n, the coefficient of xx is 3030 and the coefficient of x2x^2 is 405405.

A

Use the given coefficients to nn and kk.

I.

Form two equations involving nn and kk.

[2]
II.

nn and kk.

[4]
B

For these values of nn and kk, the coefficient of x3x^3 in (1+kx)n(1+kx)^n.

[2]
C

Let

R(x)=(1+3x)10(1−x)6R(x)=(1+3x)^{10}(1-x)^6

the coefficient of x7x^7 in R(x)R(x).

[3]
D

the largest coefficient in the expansion of (1+3x)10(1+3x)^{10}.

[2]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let

G(x)=(x2+ax)10(1+bx)5G(x)=\left(x^2+\frac{a}{x}\right)^{10}(1+bx)^5

where x≠0x\ne 0, a>0a>0 and b>0b>0. In the expansion of (x2+ax)10\left(x^2+\frac{a}{x}\right)^{10}, the coefficient of x14x^{14} is 720720.

A

aa.

I.

the general term in the expansion of (x2+ax)10\left(x^2+\frac{a}{x}\right)^{10}.

[2]
II.

aa.

[3]
B

Let b=2b=2. the constant term in the expansion of G(x)G(x). If you did not obtain a=4a=4, use a=4a=4 for this part.

[7]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^+, define

Hn(x)=(1+x)n(1+x2)nH_n(x)=(1+x)^n(1+x^2)^n

In the expansion of Hn(x)H_n(x), the coefficient of x3x^3 is 220220.

A

nn.

I.

the coefficient of x3x^3 in Hn(x)H_n(x) is (n3)+n2\binom{n}{3}+n^2.

[3]
II.

nn.

[2]
B

the coefficient of x6x^6 in H10(x)H_{10}(x).

[4]
C

Using a GDC, the largest coefficient in the expansion of H10(x)H_{10}(x).

[2]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A family of calibration polynomials is defined by A(x)=(1+kx)nA(x)=(1+kx)^n, where n∈Z+n\in\mathbb{Z}^+ and k>0k>0. The coefficient of xrx^r in the expansion of A(x)A(x) is denoted by crc_r. A partial output from a GDC for one such polynomial is shown.

rr

crc_r

0

1

1

24

2

252

3

1512

A
I.

an expression for crc_r in terms of nn, kk and rr.

[2]
II.

cr+1cr=k(n−r)r+1\dfrac{c_{r+1}}{c_r}=\dfrac{k(n-r)}{r+1}.

[3]
B

For the polynomial shown in the table, c1=24c_1=24 and c2=252c_2=252. the value of nn and the value of kk.

[4]
C

the coefficient of x4x^4 in A(x)(1−x)5A(x)(1-x)^5.

[3]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A diagonal sum in Pascal's triangle is defined by Dm=∑r=0m(r2)D_m=\sum_{r=0}^{m}\binom{r}{2}, where entries with lower index greater than upper index are taken as zero.

A
I.

D6D_6.

[1]
II.

the value of (73)\binom{7}{3} and on your answer to part (a)(i).

[2]
B
I.

(r2)+(r3)=(r+13)\binom{r}{2}+\binom{r}{3}=\binom{r+1}{3} for r≥2r\geq 2.

[2]
II.

that Dm=(m+13)D_m=\binom{m+1}{3} for all integers m≥2m\geq 2.

[2]
C

The coefficient of x2x^2 in ∑r=0m(1+x)r\sum_{r=0}^{m}(1+x)^r is 560560. mm.

[3]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A laboratory model for the remaining proportion of a substance after time tt is R(t)=(1−ct)mR(t)=(1-ct)^m, where m∈Z+m\in\mathbb{Z}^+ and 0<c<10<c<1. The first three terms in ascending powers of tt are 1−0.48t+0.1008t2+⋯1-0.48t+0.1008t^2+\cdots.

A
I.

expressions, in terms of mm and cc, for the coefficients of tt and t2t^2 in the expansion of (1−ct)m(1-ct)^m.

[2]
II.

mm and cc.

[4]
B

Using terms up to and including t3t^3, approximate R(3)R(3).

[3]
C

the percentage error in the approximation from part (b), using the exact value of R(3)R(3).

[2]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A weighted row is generated from the expansion of (1+qx)n(1+qx)^n, where n∈Z+n\in\mathbb{Z}^+ and q>0q>0. The coefficient of xrx^r is denoted by wrw_r. In one row, w2=405w_2=405 and w3w2=8\dfrac{w_3}{w_2}=8.

r

w_r

0

1

1

30

2

405

3

3240

4

17010

5

61236

6

153090

7

262440

8

295245

9

196830

10

59049

A
I.

wrw_r in terms of nn, qq and rr.

[1]
II.

w3w2=q(n−2)3\dfrac{w_3}{w_2}=\dfrac{q(n-2)}{3}.

[3]
B

nn and qq.

[4]
C

the sum of the coefficients of the odd powers of xx in (1+qx)n(1+qx)^n, and this as a coefficient sum.

[3]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
Hard
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For an even row 2m2m of Pascal's triangle, let Mm=(2mm)M_m=\binom{2m}{m} be the central coefficient.

A
I.

(2mm)(2mm−1)=m+1m\dfrac{\binom{2m}{m}}{\binom{2m}{m-1}}=\dfrac{m+1}{m}.

[2]
II.

why MmM_m is greater than each adjacent coefficient in row 2m2m.

[2]
B

Using a GDC table or otherwise, the smallest even row number 2m2m for which Mm>100000M_m>100000.

[3]
C

that (2mm)=2(2m−1m−1)\binom{2m}{m}=2\binom{2m-1}{m-1}.

[3]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

On a rectangular grid, a shortest path from (0,0)(0,0) to (p,q)(p,q) consists of pp moves right and qq moves up. Such paths can be represented by terms in the expansion of (R+U)p+q(R+U)^{p+q}.

A
I.

the number of shortest paths from (0,0)(0,0) to (7,5)(7,5).

[2]
II.

the number of shortest paths from (0,0)(0,0) to (7,5)(7,5) that pass through (3,2)(3,2).

[2]
B

the number of shortest paths from (0,0)(0,0) to (p,q)(p,q) is the coefficient of RpUqR^pU^q in (R+U)p+q(R+U)^{p+q}.

[3]
C

Vandermonde's identity ∑r=0m(kr)(n−km−r)=(nm)\sum_{r=0}^{m}\binom{k}{r}\binom{n-k}{m-r}=\binom{n}{m}, where terms with invalid lower indices are zero.

[3]
D

∑r=04(6r)(84−r)\sum_{r=0}^{4}\binom{6}{r}\binom{8}{4-r}.

[2]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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Two adjacent entries in a row of Pascal's triangle are recorded as 495495 and 792792, in that order from left to right. The row is numbered nn, starting with row 00, and the first of the two entries is (nr)\binom{n}{r}.

A
I.

(nr+1)(nr)=n−rr+1\dfrac{\binom{n}{r+1}}{\binom{n}{r}}=\dfrac{n-r}{r+1}.

[2]
II.

nn and rr.

[3]
B

the coefficient of x8x^8 in the expansion of (2+x)n(2+x)^n.

[2]
C

Using Pascal's identity, the entry directly below 495495 and 792792 in the next row is 12871287.

[3]
Question 48
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let n∈Z+n\in\mathbb{Z}^+. Consider the product

(1+x)n(1−x)n(1+x)^n(1-x)^n

In this question, take (nr)=0\binom{n}{r}=0 when r<0r<0 or r>nr>n.

A
I.

an expression, as a summation, for the coefficient of xkx^k in (1+x)n(1−x)n(1+x)^n(1-x)^n.

[3]
II.

Simplify (1+x)n(1−x)n(1+x)^n(1-x)^n.

[2]
B
I.

that, for odd kk,

∑r=0k(−1)r(nr)(nk−r)=0\sum_{r=0}^{k}(-1)^r\binom{n}{r}\binom{n}{k-r}=0
[2]
II.

that, for 0≤m≤n0\leq m\leq n,

∑r=02m(−1)r(nr)(n2m−r)=(−1)m(nm)\sum_{r=0}^{2m}(-1)^r\binom{n}{r}\binom{n}{2m-r}=(-1)^m\binom{n}{m}
[3]
C

∑r=04(−1)r(6r)(64−r)\sum_{r=0}^{4}(-1)^r\binom{6}{r}\binom{6}{4-r}
[2]

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Question 49
HL • Paper 1
Hard
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HL • Paper 1
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Consider

F(x)=(x2+1x)8(1+x)6,x≠0F(x)=\left(x^2+\frac{1}{x}\right)^8(1+x)^6,\quad x\neq 0
A
I.

Write the general term of (x2+1x)8\left(x^2+\frac{1}{x}\right)^8 corresponding to index rr.

[2]
II.

Write the condition on rr and ss for a term from this factor and a term (6s)xs\binom{6}{s}x^s from (1+x)6(1+x)^6 to contribute to the coefficient of xjx^j.

[2]
B

the constant term in the expansion of F(x)F(x).

[4]
C

the coefficient of x2x^2 in the expansion of F(x)F(x).

[3]
D

why there are no other contributions to the coefficient of x2x^2.

[1]
Question 50
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let n∈Z+n\in\mathbb{Z}^+. This question concerns sums involving the binomial coefficients in the expansion of (1+x)n(1+x)^n.

A
I.

that, for 1≤r≤n1\leq r\leq n,

r(nr)=n(n−1r−1)r\binom{n}{r}=n\binom{n-1}{r-1}
[3]
B

that

∑r=0nr(nr)=n2n−1\sum_{r=0}^{n}r\binom{n}{r}=n2^{n-1}
[3]
C

By considering the second derivative of (1+x)n(1+x)^n,

∑r=0nr(r−1)(nr)=n(n−1)2n−2\sum_{r=0}^{n}r(r-1)\binom{n}{r}=n(n-1)2^{n-2}
[3]
D

, in terms of nn,

∑r=0nr2(nr)\sum_{r=0}^{n}r^2\binom{n}{r}

and evaluate this sum when n=7n=7.

[3]
Question 51
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let p,q∈Z+p,q\in\mathbb{Z}^+ and consider the identity

(1+x)p(1+x)q=(1+x)p+q(1+x)^p(1+x)^q=(1+x)^{p+q}
A
I.

By comparing coefficients of xkx^k, that

∑r=0k(pr)(qk−r)=(p+qk)\sum_{r=0}^{k}\binom{p}{r}\binom{q}{k-r}=\binom{p+q}{k}
[5]
B

that

∑r=0n(nr)2=(2nn)\sum_{r=0}^{n}\binom{n}{r}^2=\binom{2n}{n}
[3]
C

Use this result to

∑r=06(6r)2\sum_{r=0}^{6}\binom{6}{r}^2
[4]
Question 52
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^+, consider the coefficients in the expansion of (1+x)n(1+x)^n.

A

Use coefficients to an identity.

I.

the coefficient of xnx^n in (1+x)n(1+x)n(1+x)^n(1+x)^n is ∑r=0n(nr)2\displaystyle \sum_{r=0}^{n}\binom{n}{r}^2.

[3]
II.

that

∑r=0n(nr)2=(2nn)\sum_{r=0}^{n}\binom{n}{r}^2=\binom{2n}{n}
[3]
B

Evaluate ∑r=012(12r)2\displaystyle \sum_{r=0}^{12}\binom{12}{r}^2.

[2]
C

Two subsets are chosen independently at random from a set with 1212 elements. the probability that the two subsets have the same size.

[4]
D

Using a GDC table or otherwise, the least positive integer nn for which (2nn)>1000000\binom{2n}{n}>1000000.

[3]

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Question 53
HL • Paper 2
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HL • Paper 2
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Let a>0a>0. In the expansion of (a+x)12(a+x)^{12}, the coefficients of x4x^4 and x5x^5 are equal.

A

aa.

I.

the coefficients of x4x^4 and x5x^5 in terms of aa.

[2]
II.

aa in exact form.

[2]
B

Using a=85a=\frac{8}{5}, the coefficient of x6x^6 in (a+x)12(a+x)^{12}.

[2]
C

Let

J(x)=(85+x)12(1−x)4J(x)=\left(\frac{8}{5}+x\right)^{12}(1-x)^4

the coefficient of x6x^6 in J(x)J(x).

[4]
D

For n∈Z+n\in\mathbb{Z}^+ and a>0a>0, with 0≤r≤n−10\le r\le n-1, that adjacent coefficients of xrx^r and xr+1x^{r+1} in (a+x)n(a+x)^n are equal if and only if

a=n−rr+1a=\frac{n-r}{r+1}
[3]
Question 54
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A fourth-degree binomial approximation is used for (1−3x)15(1-3x)^{15}:

B(x)=1−45x+945x2−12285x3+110565x4B(x)=1-45x+945x^2-12285x^3+110565x^4
A

the expression for B(x)B(x) using the binomial theorem.

I.

the general term in the expansion of (1−3x)15(1-3x)^{15}.

[2]
II.

the coefficient of x4x^4 in B(x)B(x) is 110565110565.

[3]
B

Use B(x)B(x) to approximate (0.94)15(0.94)^{15}.

[2]
C

Using your GDC, the percentage error in the approximation in part (b).

[3]
D

the largest value of xx in 0≤x≤0.030\leq x\leq 0.03 for which the percentage error in using B(x)B(x) to approximate (1−3x)15(1-3x)^{15} is less than 0.100%0.100\%.

[2]
Question 55
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a>0a>0 and x≠0x\neq 0, consider E(x)=(x2+ax)6(1−2x)4E(x)=\left(x^2+\dfrac{a}{x}\right)^6(1-2x)^4. Let C(a)C(a) be the constant term in the expansion of E(x)E(x).

A
I.

the power of xx in the general term formed by taking index rr from (x2+ax)6\left(x^2+\dfrac{a}{x}\right)^6 and index ss from (1−2x)4(1-2x)^4.

[2]
II.

all pairs (r,s)(r,s) which contribute to the constant term.

[1]
B

C(a)=15a4−192a5C(a)=15a^4-192a^5.

[4]
C
I.

the interval of values of aa for which C(a)>0C(a)>0.

[2]
II.

the value of aa for which C(a)C(a) is a maximum, and this maximum value.

[4]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^+, define EnE_n to be the sum of the even-indexed terms in the expansion of (1+3x)n(1+3x)^n when x=1x=1; that is, En=(n0)+(n2)32+(n4)34+⋯E_n=\binom{n}{0}+\binom{n}{2}3^2+\binom{n}{4}3^4+\cdots.

n

E_n

1

1

2

10

3

28

4

136

5

496

6

2080

A
I.

E4E_4 directly from the definition.

[2]
II.

the corresponding sum of the odd-indexed terms for n=4n=4.

[1]
B

that En=4n+(−2)n2E_n=\dfrac{4^n+(-2)^n}{2}.

[4]
C

En=524800E_n=524800 for n∈Z+n\in\mathbb{Z}^+.

[3]

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Question 57
HL • Paper 3
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HL • Paper 3
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For n∈Z+n\in\mathbb{Z}^+, define Pn(x)=(1+x)n(1+ax+bx2)P_n(x)=(1+x)^n(1+ax+bx^2). The constants aa and bb are chosen so that the coefficients of xx and x2x^2 in Pn(x)P_n(x) are both zero.

A
I.

aa in terms of nn.

[2]
II.

bb in terms of nn.

[3]
B

the coefficient of x3x^3 in Pn(x)P_n(x) is (n+23)\binom{n+2}{3}.

[3]
C

For one value of nn, the coefficient of x3x^3 is 8484. nn and the coefficient of x4x^4 in Pn(x)P_n(x).

[3]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For n∈Z+n\in\mathbb{Z}^+, define Sn=∑r=0nr(nr)2rS_n=\sum_{r=0}^{n}r\binom{n}{r}2^r. This is a weighted sum of the coefficients in the expansion of (1+2x)n(1+2x)^n when x=1x=1.

n

S_n

1

2

2

12

3

54

6

2916

5

810

A
I.

S4S_4 directly from the definition.

[2]
II.

S434\dfrac{S_4}{3^4}.

[1]
B

that Sn=2n3n−1S_n=2n3^{n-1}.

[4]
C

nn if Sn3n=18\dfrac{S_n}{3^n}=18.

[3]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A checksum uses the last two digits of powers of 99. The binomial expansion of 9n=(10−1)n9^n=(10-1)^n can be used to these last two digits for n∈Z+n\in\mathbb{Z}^+.

n

last two digits of 9^n

1

09

2

81

3

29

4

61

5

49

6

41

7

69

8

21

9

89

10

01

11

09

12

81

A
I.

Expand (10−1)n(10-1)^n using the binomial theorem, writing the first three terms.

[2]
II.

why all terms from the third term onwards do not affect the last two digits.

[2]
B

, modulo 100100, 9n≡1−10n9^n\equiv 1-10n if nn is even and 9n≡10n−19^n\equiv 10n-1 if nn is odd.

[3]
C

all values of nn with 1≤n≤501\leq n\leq 50 for which the last two digits of 9n9^n are 2121.

[3]
Question 60
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For n∈Z+n\in\mathbb{Z}^+, write

5n=(1+4)n5^n=(1+4)^n

An integer is said to be divisible by 1616 if it can be written in the form 16m16m for some integer mm.

A
I.

Write the first three terms in the binomial expansion of (1+4)n(1+4)^n.

[3]
II.

5n5^n can be written in the form

5n=1+4n+16M5^n=1+4n+16M

where MM is an integer.

[2]
B
I.

that 5n−15^n-1 is divisible by 1616 if nn is divisible by 44.

[2]
II.

the converse: if 5n−15^n-1 is divisible by 1616, then nn is divisible by 44.

[2]
C

whether 52028−15^{2028}-1 is divisible by 1616.

[2]
D

the remainder when 520275^{2027} is divided by 1616.

[2]

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Complex Numbers