Date | November 2016 | Marks available | 6 | Reference code | 16N.3ca.hl.TZ0.2 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
By successive differentiation find the first four non-zero terms in the Maclaurin series for f(x)=(x+1)ln(1+x)−x.
Deduce that, for n⩾2, the coefficient of xn in this series is (−1)n1n(n−1).
By applying the ratio test, find the radius of convergence for this Maclaurin series.
Markscheme
f(x)=(x+1)ln(1+x)−x f(0)=0 A1
f′(x)=ln(1+x)+x+11+x−1 (=ln(1+x)) f′(0)=0 M1A1A1
f″(x)=(1+x)−1 f″(0)=1 A1A1
f‴(x)=−(1+x)−2 f‴(0)=−1 A1
f(4)(x)=2(1+x)−3 f(4)(0)=2 A1
f(5)(x)=−3×2(1+x)−4 f(5)(0)=−3×2 A1
f(x)=x22!−1x33!+2x44!−6x55!… M1A1
f(x)=x21×2−x32×3+x43×4−x54×5…
f(x)=x22−x36+x412−x520…
Note: Allow follow through from the first error in a derivative (provided future derivatives also include the chain rule), no follow through after a second error in a derivative.
[11 marks]
f(n)(0)=(−1)n(n−2)! So coefficient of xn=(−1)n(n−2)!n! A1
coefficient of xn is (−1)n1n(n−1) AG
[1 mark]
applying the ratio test to the series of absolute terms
lim M1A1
= \mathop {\lim }\limits_{n \to \infty } \left| x \right|\frac{{(n - 1)}}{{(n + 1)}} A1
= \left| x \right| A1
so for convergence \left| x \right| < 1, giving radius of convergence as 1 (M1)A1
[6 marks]