CAIE IGCSE Additional Math Combinatorics Series and Vectors Questions

Use this algebraic-methods unit hub to connect restricted counting with binomial and progression methods, then apply vectors to geometry and motion.

Exam points

Solve permutation, combination and restricted-counting problems using notation, cases, blocks or complements.
Expand binomials, select coefficients and solve arithmetic or geometric progression conditions.
Use component, position and velocity vectors for ratios, intersections, distances and collision or meeting tests.

Covered syllabus points

11. Permutations and combinations 12. Series 13. Vectors in two dimensions

[Maximum number: 9]

(a)

Find the first three terms in the expansion of $\left(1+\frac{x}{7}\right)^5$ , in ascending powers of x. Simplify the coefficient of each term.

[ 2 ]

$1+\frac57x+\frac{10}{49}x^2$
B1 for any two correct terms or for the three terms listed but not summed

(b)

The expansion of $7(1+x)^n\left(1+\frac{x}{7}\right)^5$ , where n is a positive integer, is written in ascending powers of x. The first two terms in the expansion are 7+89x. Find the value of n.

[ 2 ]

12
B1 for 7n+5=89 or $7(n+\frac57)=89$ oe or ${}^nC_1=12$

(c)

In the expansion of $(k-2x)^8$ , where k is a constant, the coefficient of $x^4$ divided by the coefficient of $x^2$ is $\frac{5}{8}$ . The coefficient of x is positive. Form an equation and hence find the value of k.

[ 5 ]

${}^8C_4\times k^4\times(-2)^4[x^4]$
oe, or $1120k^4[x^4]$ soi
${}^8C_2\times k^6\times(-2)^2[x^2]$
oe, or $112k^6[x^2]$ soi
$\frac{1120k^4}{112k^6}=\frac58$
oe, soi
M1 FT providing at least B1 awarded and correct terms attempted
$k^2=16$
soi
For coefficient of x to be positive, k<0, therefore k=-4

[Maximum number: 3]

the equation $(n-4)^{n+1} C_{5}={ }^{n+2} C_{7}$ .

$\frac{(n-4)(n+1)!}{(n-4)!5!}=\frac{(n+2)!}{(n-5)!7!}$

leading to
$\times 6=n+2$ B1 for $7 \times 6$ or 42 seen as a result of simplification of the numerical factorials

B1 for n+2 seen as a result of simplification of the algebraic factorials
n=40 B1

[Maximum number: 15]

An arithmetic progression, A, has first term a and common difference d. The 2nd, 14th and 17th terms of A form the first three terms of a convergent geometric progression, G, with common ratio r .

(a)

Given that $d\neq0$ , find two expressions for r in terms of a and d and hence show that a=-17d.

[ 6 ]

$a+d,\quad a+13d,\quad a+16d$
soi
$\frac{a+13d}{a+d}=\frac{a+16d}{a+13d}$
oe
$a^2+26ad+169d^2=a^2+17ad+16d^2$ $9ad+153d^2=0$ $9d(a+17d)=0\Rightarrow a=-17d$
oe

(b)

Find the value of r .

[ 2 ]

r=0.25
oe
M1 for $\frac{-17d+13d}{-17d+d}$ or $\frac{-17d+16d}{-17d+13d}$

(c)

The first term of the geometric progression, G , is q and the sum to infinity is $\frac{256}{3}$ . Find the sum of the first 20 terms of the arithmetic progression, A .

[ 7 ]

$\frac{q}{1-0.25}=\frac{256}{3}$
oe
q=64-17d+d=64
oe
$d=-4,\quad a=68$ $S_{20}=600$

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question 3

Question (a)

Question (b)

Question (c)