Date | May 2016 | Marks available | 4 | Reference code | 16M.2.sl.TZ1.4 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
Find the term in x6 in the expansion of (x+2)9.
Hence, find the term in x7 in the expansion of 5x(x+2)9.
Markscheme
valid approach to find the required term (M1)
eg(9r)(x)9−r(2)r, x9+9x8(2)+(92)x7(2)2+…, Pascal’s triangle to the 9th row
identifying correct term (may be indicated in expansion) (A1)
eg4th term, r=6, (93), (x)6(2)3
correct calculation (may be seen in expansion) (A1)
eg(93)(x)6(2)3, 84×23
672x6 A1 N3
[4 marks]
valid approach (M1)
egrecognizing x7 is found when multiplying 5x×672x6
3360x7 A1 N2
[2 marks]
Examiners report
Many candidates approached this question using an appropriate and efficient method to identify the required term. While many of those who were successful in (a) were also successful in (b), a significant number of candidates did not realize that multiplying their answer in (a) by 5x would give them the term in x7. This led to an attempt to find a binomial expansion, which was generally unsuccessful. Some candidates continue to be unable to distinguish between the “coefficient” and the “term” and lost a point as a result.
Many candidates approached this question using an appropriate and efficient method to identify the required term. While many of those who were successful in (a) were also successful in (b), a significant number of candidates did not realize that multiplying their answer in (a) by 5x would give them the term in x7. This led to an attempt to find a binomial expansion, which was generally unsuccessful. Some candidates continue to be unable to distinguish between the “coefficient” and the “term” and lost a point as a result.