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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 $x$ $x$ is defined as a function of the form $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . .$ $f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$ where the ${a_i} \in \mathbb{R}$ .

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 ${\left( {1 + x} \right)^q}{\text{,}}\,\,q \in \mathbb{Q}$ can be written as the power series ${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$ .

Expand ${\left( {1 + x} \right)^5}$ using the Binomial Theorem.

[2]

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 ${\left( {1 + x} \right)^{ - 1}}$ and state the values of $x$ for which this equality is true.

[4]

Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for ${\left( {1 + x} \right)^{ - 2}}$ .

[2]

Repeat this process to find the first four terms in a power series for ${\left( {1 + x} \right)^{ - 3}}$ .

[2]

Hence, by recognising the pattern, deduce the first four terms in a power series for ${\left( {1 + x} \right)^{ - n}}$ , $n \in {\mathbb{Z}^ + }$ .

[3]

By substituting $x = 0$ , find the value of ${a_0}$ .

[1]

By differentiating both sides of the expression and then substituting $x = 0$ , find the value of ${a_1}$ .

[2]

Repeat this procedure to find ${a_2}$ and ${a_3}$ .

[4]

Hence, write down the first four terms in what is called the Extended Binomial Theorem for ${\left( {1 + x} \right)^q}{\text{,}}\,\,q \in \mathbb{Q}$ .

[1]

Write down the power series for $\frac{1}{{1 + {x^2}}}$ .

[2]

Hence, using integration, find the power series for ${\text{arctan}}\,x$ , giving the first four non-zero terms.

[4]

Markscheme

$1 + 5x + 10{x^2} + 10{x^3} + 5{x^4} + {x^5}$ M1A1

[2 marks]

It is an infinite GP with $a = 1{\text{,}}\,\,r = - x$ R1A1

${S_\infty } = \frac{1}{{1 - \left( { - x} \right)}} = \frac{1}{{1 + x}} = {\left( {1 + x} \right)^{ - 1}}$ M1A1AG

[4 marks]

${\left( {1 + x} \right)^{ - 1}} = 1 - x + {x^2} - {x^3} + {x^4} - ...$

$- 1{\left( {1 + x} \right)^{ - 2}} = - 1 + 2x - 3{x^2} + 4{x^3} - ...$ A1

${\left( {1 + x} \right)^{ - 2}} = 1 - 2x + 3{x^2} - 4{x^3} + ...$ A1

[2 marks]

$- 2{\left( {1 + x} \right)^{ - 3}} = - 2 + 6x - 12{x^2} + 20{x^3}...$ A1

${\left( {1 + x} \right)^{ - 3}} = 1 - 3x + 6{x^2} - 10{x^3}...$ A1

[2 marks]

${\left( {1 + x} \right)^{ - n}} = 1 - nx + \frac{{n\left( {n + 1} \right)}}{{2{\text{!}}}}{x^2} - \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3{\text{!}}}}{x^3}...$ A1A1A1

[3 marks]

${1^q} = {a_0} \Rightarrow {a_0} = 1$ A1

[1 mark]

$q{\left( {1 + x} \right)^{q - 1}} = {a_1} + 2{a_2}x + 3{a_3}{x^2} + ...$ A1

${a_1} = q$ A1

[2 marks]

$q\left( {q - 1} \right){\left( {1 + x} \right)^{q - 2}} = 1 \times 2{a_2} + 2 \times 3{a_3}x + ...$ A1

${a_2} = \frac{{q\left( {q - 1} \right)}}{{2{\text{!}}}}$ A1

$q\left( {q - 1} \right)\left( {q - 2} \right){\left( {1 + x} \right)^{q - 3}} = 1 \times 2 \times 3{a_3} + ...$ A1

${a_3} = \frac{{q\left( {q - 1} \right)\left( {q - 2} \right)}}{{3{\text{!}}}}$ A1

[4 marks]

${\left( {1 + x} \right)^q} = 1 + qx + \frac{{q\left( {q - 1} \right)}}{{2{\text{!}}}}{x^2} + \frac{{q\left( {q - 1} \right)\left( {q - 2} \right)}}{{3{\text{!}}}}{x^3}...$ A1