CAIE IGCSE Additional Math 12 series

Use this Series hub to move from binomial expansions into finite arithmetic and geometric series, convergence and sum-to-infinity problems.

Covered syllabus points

12.1. Use the binomial theorem 12.2. Use the general binomial term 12.3. Recognise arithmetic and geometric progressions 12.4. Use finite progression formulas 12.5. Use convergence and sum to infinity

[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 ]

$1+\frac57x+\frac{10}{49}x^2$
B2
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
B2
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
B1
${}^8C_2\times k^6\times(-2)^2[x^2]$
oe, or $112k^6[x^2]$ soi
B1
$\frac{1120k^4}{112k^6}=\frac58$
oe, soi
M1 FT providing at least B1 awarded and correct terms attempted
$k^2=16$
soi
A1
For coefficient of x to be positive, k<0, therefore k=-4
A1

[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
B1
$\frac{a+13d}{a+d}=\frac{a+16d}{a+13d}$
oe
M2
$a^2+26ad+169d^2=a^2+17ad+16d^2$
A1
$9ad+153d^2=0$
A1
$9d(a+17d)=0\Rightarrow a=-17d$
oe
A1

(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
M1
q=64
A1
-17d+d=64
oe
DM1
$d=-4,\quad a=68$
A2
$S_{20}=600$
A1

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Covered syllabus points

12. Series question 1

Question (a)

Question (b)

Question (c)

12. Series question 2

Question (a)

Question (b)

Question (c)