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Binomial Theorem

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

Verified by Karim
Verified by Karim
Paper
Difficulty
Status
Level
Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

Consider the expansion of (23x)5(2-3x)^5 in ascending powers of xx.

A

Find the coefficient of x2x^2.

[2]
B

Find 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

Show that k=2k=2.

[3]
B

Hence find the coefficient of x3x^3.

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

Consider the expansion of (23x)8(2-3x)^8.

A

Write down the number of terms in the expansion.

[1]
B

Find 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

Write down the value of nn.

[1]
B

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

[2]
C

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

[1]
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 nZ+n\in\mathbb{Z}^+ and n3n\geq 3.

A

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

[3]
B

Find 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 x0x\neq 0.

A

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

[2]
B

Find 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 pZ+p\in\mathbb{Z}^+ and p>1p>1.

A

Find the possible values of pp.

[3]
B

Hence find 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

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

[2]
B

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

[2]
C

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

[1]
Question 9
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Consider the expansion of

(x2+3x)9,x0\left(x^2+\frac{3}{x}\right)^9, \quad x\ne 0
A

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

[2]
B

Find 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, determine all values of rr for which (15r)>3000\binom{15}{r}>3000.

[2]
B

Hence determine 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

Find aa.

[2]
B

Find bb.

[2]
C

Using your value of aa, find the coefficient of x2x^2 in (1+ax)8(1x)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 (3x2)7(3x-2)^7, the terms are written in descending powers of xx.

A

Find the fourth term.

[3]
B

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

[3]
Question 13
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

In the expansion of (1+mx)7(12x)(1+mx)^7(1-2x), the coefficient of xx is 55.

A

Show that m=1m=1.

[3]
B

Hence find the coefficient of x3x^3 in the expansion of (1+mx)7(12x)(1+mx)^7(1-2x).

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

Consider the expansion of

(2x21x)8\left(2x^2-\frac{1}{x}\right)^8

where x0x\neq 0.

A

Find the general term in the expansion corresponding to rr.

[2]
B

Find 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

Write down 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, calculate the percentage error in this approximation.

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

Let nNn\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

Hence find nn.

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

Consider the polynomial

(1+2x)8(1x)5(1+2x)^8(1-x)^5
A

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

[1]
B

Determine 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

(ax23x)9,x0\left(ax^2-\frac{3}{x}\right)^9, \quad x\ne 0

the constant term is 489888489888.

A

Show that the constant term is 61236a361236a^3.

[3]
B

Find 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 nNn\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

Hence determine 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

Find the value of nn.

[3]
B

Hence find the coefficient of x5x^5.

[2]
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(1qx)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.

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

[2]
II.

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

[3]
B

Determine the values of pp and qq.

[4]
C

Hence find 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.

Find the value of aa.

[2]
II.

Find the value of bb.

[2]
B

Verify that the coefficient of x2x^2 is 672672.

[2]
C

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

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

For nZ+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.

Write down an expression for En+OnE_n+O_n.

[1]
II.

Find an expression for EnOnE_n-O_n.

[2]
B

Hence prove that

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

Find the value of E6E_6.

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

Let nZ+n\in\mathbb{Z}^+ with n3n\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.

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

[2]
II.

Find the value of nn.

[3]
B

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

(1+x)n(1x)2(1+x)^n(1-x)^2
[4]
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 nZ+n\in\mathbb{Z}^+.

A

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

[2]
B

Determine 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

Find aa.

[2]
B

Find bb.

[3]
C

Verify 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 nZ+n\in\mathbb{Z}^+.

A

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

[2]
B

Using (11)n(1-1)^n, prove that S=2n1S=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.

Write down 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, calculate 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. Determine the largest value of xx, where 0x0.050\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]
Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Let

P(x)=(2+px)7(1qx)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 find pp and qq.

I.

Show that 7p6q=17p-6q=1.

[3]
II.

Hence find pp and qq.

[3]
B

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

[3]
C

For these values of pp and qq, find 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 x0x\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 0r80\leq r\leq 8.

I.

Write down the general term.

[2]
II.

Hence state all possible exponents of xx in the expansion.

[2]
B

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

[3]
C

Using your value of aa, find the coefficient of x2x^{-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. Show that the row number is 2323.

[3]
II.

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

[1]
B

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

[3]
C

Use a counting argument to justify Pascal's identity

(n+1r)=(nr1)+(nr)\binom{n+1}{r}=\binom{n}{r-1}+\binom{n}{r}

where 1rn1\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 312r2r3^{12-r}2^r.

A simple grid-route diagram showing a start point, an end point after 12 total moves, and examples of east and north moves labelled with weights 3 and 2 respectively.
A

Use binomial coefficients to count routes.

I.

Write down the number of routes with exactly rr north moves.

[1]
II.

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

[2]
B

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

[2]
C

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

[3]
Question 33
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the product

(x+2x2)7(1x)8,x0\left(x+\frac{2}{x^2}\right)^7(1-x)^8, \quad x\ne 0
A

Find 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 (1x)8(1-x)^8.

[2]
B

Determine 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(1x)4,x0E(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.

Write down 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.

Show that 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 (1x)4(1-x)^4.

[2]
B

Find the value of kk.

[3]
C

Hence find 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.

Show 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

Find the value of abab.

[2]
C

Determine the possible values of aa and bb.

[2]
D

Find 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 nZ+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

Find the values of nn and kk.

[4]
C

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

[2]
D

Hence find the coefficient of x3x^3 in

(1+kx)n(1x)4(1+kx)^n(1-x)^4
[2]
Question 37
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

Let nZ+n\in\mathbb{Z}^+, n3n\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.

Write down expressions for these two coefficients.

[2]
II.

Hence determine nn.

[4]
B

Let

Q(x)=(1+x)8(12x)4Q(x)=(1+x)^8(1-2x)^4

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

[4]
C

Using a GDC, determine the greatest magnitude of any coefficient in the expansion of (1+x)8(12x)4(1+x)^8(1-2x)^4.

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

Let nZ+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 find nn and kk.

I.

Form two equations involving nn and kk.

[2]
II.

Hence determine nn and kk.

[4]
B

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

[2]
C

Let

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

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

[3]
D

Determine 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 x0x\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

Find aa.

I.

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

[2]
II.

Hence find aa.

[3]
B

Let b=2b=2. Find 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 nZ+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

Find nn.

I.

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

[3]
II.

Hence find nn.

[2]
B

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

[4]
C

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

[2]
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 nZ+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.

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

[2]
II.

Show that cr+1cr=k(nr)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. Determine the value of nn and the value of kk.

[4]
C

Hence find the coefficient of x4x^4 in A(x)(1x)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 portion of Pascal's triangle with one shallow diagonal corresponding to entries of the form binomial r choose 2 highlighted, and the adjacent entry at the end of the diagonal indicated but not numerically evaluated.
A
I.

Calculate D6D_6.

[1]
II.

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

[2]
B
I.

Show that (r2)+(r3)=(r+13)\binom{r}{2}+\binom{r}{3}=\binom{r+1}{3} for r2r\geq 2.

[2]
II.

Prove that Dm=(m+13)D_m=\binom{m+1}{3} for all integers m2m\geq 2.

[2]
C

The coefficient of x2x^2 in r=0m(1+x)r\sum_{r=0}^{m}(1+x)^r is 560560. Hence determine 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)=(1ct)mR(t)=(1-ct)^m, where mZ+m\in\mathbb{Z}^+ and 0<c<10<c<1. The first three terms in ascending powers of tt are 10.48t+0.1008t2+1-0.48t+0.1008t^2+\cdots.

Exact and cubic approximation of the remaining proportion.
A
I.

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

[2]
II.

Find mm and cc.

[4]
B

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

[3]
C

Calculate 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 nZ+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.

Write down wrw_r in terms of nn, qq and rr.

[1]
II.

Show that w3w2=q(n2)3\dfrac{w_3}{w_2}=\dfrac{q(n-2)}{3}.

[3]
B

Determine nn and qq.

[4]
C

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

[3]
Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

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

Rows of Pascal's triangle near an even-numbered row, with the central entry of the even row highlighted and the two adjacent entries indicated.
A
I.

Show that (2mm)(2mm1)=m+1m\dfrac{\binom{2m}{m}}{\binom{2m}{m-1}}=\dfrac{m+1}{m}.

[2]
II.

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

[2]
B

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

[3]
C

Prove that (2mm)=2(2m1m1)\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 rectangular grid from $(0,0)$ to $(7,5)$ with a marked intermediate point $(3,2)$. Right and up directions are labelled, and no path counts are displayed.
A
I.

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

[2]
II.

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

[2]
B

Show that 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

Prove Vandermonde's identity r=0m(kr)(nkmr)=(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

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

[2]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

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 segment of one row of Pascal's triangle showing two adjacent boxed entries labelled 495 and 792, with row number $n$ and position $r$ indicated symbolically.
A
I.

Show that (nr+1)(nr)=nrr+1\dfrac{\binom{n}{r+1}}{\binom{n}{r}}=\dfrac{n-r}{r+1}.

[2]
II.

Determine nn and rr.

[3]
B

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

[2]
C

Using Pascal's identity, show that 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 nZ+n\in\mathbb{Z}^+. Consider the product

(1+x)n(1x)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.

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

[3]
II.

Simplify (1+x)n(1x)n(1+x)^n(1-x)^n.

[2]
B
I.

Prove that, for odd kk,

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

Prove that, for 0mn0\leq m\leq n,

r=02m(1)r(nr)(n2mr)=(1)m(nm)\sum_{r=0}^{2m}(-1)^r\binom{n}{r}\binom{n}{2m-r}=(-1)^m\binom{n}{m}
[3]
C

Hence find

r=04(1)r(6r)(64r)\sum_{r=0}^{4}(-1)^r\binom{6}{r}\binom{6}{4-r}
[2]
Question 49
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Consider

F(x)=(x2+1x)8(1+x)6,x0F(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

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

[4]
C

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

[3]
D

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

A
I.

Prove that, for 1rn1\leq r\leq n,

r(nr)=n(n1r1)r\binom{n}{r}=n\binom{n-1}{r-1}
[3]
B

Deduce that

r=0nr(nr)=n2n1\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, show that

r=0nr(r1)(nr)=n(n1)2n2\sum_{r=0}^{n}r(r-1)\binom{n}{r}=n(n-1)2^{n-2}
[3]
D

Hence find, 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,qZ+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, prove that

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

Hence prove that

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

Use this result to find

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

A

Use coefficients to prove an identity.

I.

Show that 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.

Hence prove 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. Find the probability that the two subsets have the same size.

[4]
D

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

[3]
Question 53
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

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

Find aa.

I.

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

[2]
II.

Hence find aa in exact form.

[2]
B

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

[2]
C

Let

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

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

[4]
D

For nZ+n\in\mathbb{Z}^+ and a>0a>0, with 0rn10\le r\le n-1, justify 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=nrr+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 (13x)15(1-3x)^{15}:

B(x)=145x+945x212285x3+110565x4B(x)=1-45x+945x^2-12285x^3+110565x^4
Exact and 4th-degree curves for (1-3x)^15 on 0≤x≤0.03.
A

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

I.

Write down the general term in the expansion of (13x)15(1-3x)^{15}.

[2]
II.

Show that 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, calculate the percentage error in the approximation in part (b).

[3]
D

Determine the largest value of xx in 0x0.030\leq x\leq 0.03 for which the percentage error in using B(x)B(x) to approximate (13x)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 x0x\neq 0, consider E(x)=(x2+ax)6(12x)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).

Graph of the constant term function C(a) for a>0.
A
I.

Find 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 (12x)4(1-2x)^4.

[2]
II.

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

[1]
B

Show that C(a)=15a4192a5C(a)=15a^4-192a^5.

[4]
C
I.

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

[2]
II.

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

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

For nZ+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.

Calculate E4E_4 directly from the definition.

[2]
II.

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

[1]
B

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

[4]
C

Hence solve En=524800E_n=524800 for nZ+n\in\mathbb{Z}^+.

[3]
Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For nZ+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.

Find aa in terms of nn.

[2]
II.

Find bb in terms of nn.

[3]
B

Show that 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. Hence determine nn and find 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 nZ+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.

Calculate S4S_4 directly from the definition.

[2]
II.

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

[1]
B

Prove that Sn=2n3n1S_n=2n3^{n-1}.

[4]
C

Hence find 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=(101)n9^n=(10-1)^n can be used to investigate these last two digits for nZ+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 (101)n(10-1)^n using the binomial theorem, writing the first three terms.

[2]
II.

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

[2]
B

Show that, modulo 100100, 9n110n9^n\equiv 1-10n if nn is even and 9n10n19^n\equiv 10n-1 if nn is odd.

[3]
C

Determine all values of nn with 1n501\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 nZ+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.

Show that 5n5^n can be written in the form

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

where MM is an integer.

[2]
B
I.

Prove that 5n15^n-1 is divisible by 1616 if nn is divisible by 44.

[2]
II.

Prove the converse: if 5n15^n-1 is divisible by 1616, then nn is divisible by 44.

[2]
C

Hence determine whether 5202815^{2028}-1 is divisible by 1616.

[2]
D

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

[2]

Complex Numbers