Download IITian Academy App for accessing Online Mock Tests and More..
Question
The function \(f(x) = \frac{{1 + ax}}{{1 + bx}}\) can be expanded as a power series in x, within its radius of convergence R, in the form \(f(x) \equiv 1 + \sum\limits_{n = 1}^\infty {{c_n}{x^n}} \) .
(a) (i) Show that \({c_n} = {( – b)^{n – 1}}(a – b)\).
(ii) State the value of R.
(b) Determine the values of a and b for which the expansion of f(x) agrees with that of \({{\text{e}}^x}\) up to and including the term in \({x^2}\) .
(c) Hence find a rational approximation to \({{\text{e}}^{\frac{1}{3}}}\) .
▶️Answer/Explanation
Markscheme
(a) (i) \(f(x) = (1 + ax){(1 + bx)^{ – 1}}\)
\( = (1 + ax)(1 – bx + …{( – 1)^n}{b^n}{x^n} + …\) M1A1
it follows that
\({c_n} = {( – 1)^n}{b^n} + {( – 1)^{n – 1}}a{b^{n – 1}}\) M1A1
\( = {( – b)^{n – 1}}(a – b)\) AG
(ii) \({\text{R}} = \frac{1}{{\left| b \right|}}\) A1
[5 marks]
(b) to agree up to quadratic terms requires
\(1 = – b + a,{\text{ }}\frac{1}{2} = {b^2} – ab\) M1A1A1
from which \(a = – b = \frac{1}{2}\) A1
[4 marks]
(c) \({{\text{e}}^x} \approx \frac{{1 + 0.5x}}{{1 – 0.5x}}\) A1
putting \(x = \frac{1}{3}\) M1
\({{\text{e}}^{\frac{1}{3}}} \approx \frac{{\left( {1 + \frac{1}{6}} \right)}}{{\left( {1 – \frac{1}{6}} \right)}} = \frac{7}{5}\) A1
[3 marks]
Total [12 marks]
Examiners report
Most candidates failed to realize that the first step was to write f(x) as \((1 + ax){(1 + bx)^{ – 1}}\) . Given the displayed answer to part(a), many candidates successfully tackled part(b). Few understood the meaning of the ‘hence’ in part(c).
Question
a.Find the radius of convergence of the infinite series
\[\frac{1}{2}x + \frac{{1 \times 3}}{{2 \times 5}}{x^2} + \frac{{1 \times 3 \times 5}}{{2 \times 5 \times 8}}{x^3} + \frac{{1 \times 3 \times 5 \times 7}}{{2 \times 5 \times 8 \times 11}}{x^4} + \ldots {\text{ .}}\][7]
b.Determine whether the series \(\sum\limits_{n = 1}^\infty {\sin \left( {\frac{1}{n} + n\pi } \right)} \) is convergent or divergent.[8]
▶️Answer/Explanation
Markscheme
the nth term is
\({u_n} = \frac{{1 \times 3 \times 5 \ldots (2n – 1)}}{{2 \times 5 \times 8 \ldots (3n – 1)}}{x^n}\) M1A1
(using the ratio test to test for absolute convergence)
\(\frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{(2n + 1)}}{{(3n + 2)}}\left| x \right|\) M1A1
\(\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{u_{n + 1}}} \right|}}{{\left| {{u_n}} \right|}} = \frac{{2\left| x \right|}}{3}\) A1
let R denote the radius of convergence
then \(\frac{{2R}}{3} = 1\) so \(r = \frac{3}{2}\) M1A1
Note: Do not penalise the absence of absolute value signs.
[7 marks]
using the compound angle formula or a graphical method the series can be written in the form (M1)
\(\sum\limits_{n = 1}^\infty {{u_n}} \) where \({u_n} = {( – 1)^n}\sin \left( {\frac{1}{n}} \right)\) A2
since \(\frac{1}{n} < \frac{\pi }{2}\) i.e. an angle in the first quadrant, R1
it is an alternating series R1
\({u_n} \to 0{\text{ as }}n \to \infty \) R1
and \(\left| {{u_{n + 1}}} \right| < \left| {{u_n}} \right|\) R1
it follows that the series is convergent R1
[8 marks]
Examiners report
Solutions to this question were generally disappointing. In (a), many candidates were unable even to find an expression for the nth term so that they could not apply the ratio test.
Solutions to this question were generally disappointing. In (b), few candidates were able to rewrite the nth term in the form \(\sum {{{( – 1)}^n}\sin \left( {\frac{1}{n}} \right)} \) so that most candidates failed to realise that the series was alternating.
| Marks available | 9 |
| Reference code | 13N.3ca.hl.TZ0.5 |
Question
Each term of the power series \(\frac{1}{{1 \times 2}} + \frac{1}{{4 \times 5}}x + \frac{1}{{7 \times 8}}{x^2} + \frac{1}{{10 \times 11}}{x^3} + \ldots \) has the form \(\frac{1}{{b(n) \times c(n)}}{x^n}\), where \(b(n)\) and \(c(n)\) are linear functions of \(n\).
(a) Find the functions \(b(n)\) and \(c(n)\).
(b) Find the radius of convergence.
(c) Find the interval of convergence.
▶️Answer/Explanation
Markscheme
(a) \(b(n) = 3n + 1\) A1
\(c(n) = 3n + 2\) A1
Note: \(b(n)\) and \(c(n)\) may be reversed.
[2 marks]
(b) consider the ratio of the \({(n + 1)^{{\text{th}}}}\) and \({n^{{\text{th}}}}\) terms: M1
\(\frac{{3n + 1}}{{3n + 4}} \times \frac{{3n + 2}}{{3n + 5}} \times \frac{{{x^{n + 1}}}}{{{x^n}}}\) A1
\(\mathop {{\text{lim}}}\limits_{n \to 0} \frac{{3n + 1}}{{3n + 4}} \times \frac{{3n + 2}}{{3n + 5}} \times \frac{{{x^{n + 1}}}}{{{x^n}}}x\) A1
radius of convergence: \(R = 1\) A1
[4 marks]
(c) any attempt to study the series for \(x = -1\) or \(x = 1\) (M1)
converges for \(x = 1\) by comparing with p-series \(\sum {\frac{1}{{{n^2}}}} \) R1
attempt to use the alternating series test for \(x = -1\) (M1)
Note: At least one of the conditions below needs to be attempted for M1.
\(\left| {{\text{terms}}} \right| \approx \frac{1}{{9{n^2}}} \to 0\) and terms decrease monotonically in absolute value A1
series converges for \(x = -1\) R1
interval of convergence: \(\left[ { – 1,{\text{ 1}}} \right]\) A1
Note: Award the R1s only if an attempt to corresponding correct test is made;
award the final A1 only if at least one of the R1s is awarded;
Accept study of absolute convergence at end points.
[6 marks]