User interface language: English | Español

Date May Example questions Marks available 3 Reference code EXM.3.AHL.TZ0.1
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Hence and Deduce Question number 1 Adapted from N/A

Question

This question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.

A power series in x is defined as a function of the form  f ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + . . . where the a i R .

It can be considered as an infinite polynomial.

This is an example of a power series, but is only a finite power series, since only a finite number of the a i are non-zero.

We will now attempt to generalise further.

Suppose  ( 1 + x ) q , q Q  can be written as the power series  a 0 + a 1 x + a 2 x 2 + a 3 x 3 + . . . .

Expand  ( 1 + x ) 5  using the Binomial Theorem.

[2]
a.

Consider the power series  1 x + x 2 x 3 + x 4 . . .

By considering the ratio of consecutive terms, explain why this series is equal to  ( 1 + x ) 1 and state the values of x for which this equality is true.

[4]
b.

Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for  ( 1 + x ) 2 .

[2]
c.

Repeat this process to find the first four terms in a power series for ( 1 + x ) 3 .

[2]
d.

Hence, by recognising the pattern, deduce the first four terms in a power series for ( 1 + x ) n , n Z + .

[3]
e.

By substituting x = 0 , find the value of a 0 .

[1]
f.

By differentiating both sides of the expression and then substituting x = 0 , find the value of a 1 .

[2]
g.

Repeat this procedure to find a 2 and a 3 .

[4]
h.

Hence, write down the first four terms in what is called the Extended Binomial Theorem for  ( 1 + x ) q , q Q .

[1]
i.

Write down the power series for 1 1 + x 2 .

[2]
j.

Hence, using integration, find the power series for arctan x , giving the first four non-zero terms.

[4]
k.

Markscheme

1 + 5 x + 10 x 2 + 10 x 3 + 5 x 4 + x 5       M1A1

[2 marks]

a.

It is an infinite GP with  a = 1 , r = x       R1A1

S = 1 1 ( x ) = 1 1 + x = ( 1 + x ) 1       M1A1AG

[4 marks]

b.

( 1 + x ) 1 = 1 x + x 2 x 3 + x 4 . . .

1 ( 1 + x ) 2 = 1 + 2 x 3 x 2 + 4 x 3 . . .        A1

( 1 + x ) 2 = 1 2 x + 3 x 2 4 x 3 + . . .        A1

 

[2 marks]

c.

2 ( 1 + x ) 3 = 2 + 6 x 12 x 2 + 20 x 3 . . .       A1

( 1 + x ) 3 = 1 3 x + 6 x 2 10 x 3 . . .       A1

[2 marks]

d.

( 1 + x ) n = 1 n x + n ( n + 1 ) 2 ! x 2 n ( n + 1 ) ( n + 2 ) 3 ! x 3 . . .      A1A1A1

[3 marks]

e.

1 q = a 0 a 0 = 1       A1

[1 mark]

f.

q ( 1 + x ) q 1 = a 1 + 2 a 2 x + 3 a 3 x 2 + . . .        A1

a 1 = q        A1

[2 marks]

g.

q ( q 1 ) ( 1 + x ) q 2 = 1 × 2 a 2 + 2 × 3 a 3 x + . . .        A1

a 2 = q ( q 1 ) 2 !        A1

q ( q 1 ) ( q 2 ) ( 1 + x ) q 3 = 1 × 2 × 3 a 3 + . . .        A1

a 3 = q ( q 1 ) ( q 2 ) 3 !        A1

[4 marks]

h.

( 1 + x ) q = 1 + q x + q ( q 1 ) 2 ! x 2 + q ( q 1 ) ( q 2 ) 3 ! x 3 . . .      A1

[1 mark]

i.

1 1 + x 2 = 1 x 2 + x 4 x 6 + . . .     M1A1

[2 marks]

j.

arctan x + c = x x 3 3 + x 5 5 x 7 7 + . . .     M1A1

Putting  x = 0 c = 0         R1

So  arctan x = x x 3 3 + x 5 5 x 7 7 + . . .         A1

[4 marks]

k.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.
[N/A]
g.
[N/A]
h.
[N/A]
i.
[N/A]
j.
[N/A]
k.

Syllabus sections

Topic 1—Number and algebra » AHL 1.10—Perms and combs, binomial with negative and fractional indices
Show 42 related questions
Topic 5 —Calculus » AHL 5.19—Maclaurin series
Topic 1—Number and algebra
Topic 5 —Calculus

View options