Date | May 2015 | Marks available | 4 | Reference code | 15M.2.sl.TZ1.2 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
Consider the expansion of (2x+3)8.
Write down the number of terms in this expansion.
Find the term in x3.
Markscheme
9 terms A1 N1
[1 mark]
valid approach to find the required term (M1)
eg(8r)(2x)8−r(3)r, (2x)8(3)0+(2x)7(3)1+…, Pascal’s triangle to 8th row
identifying correct term (may be indicated in expansion) (A1)
eg6th term, r=5, (85), (2x)3(3)5
correct working (may be seen in expansion) (A1)
eg(85)(2x)3(3)5, 56×23×35
108864x3(accept 109000x3) A1 N3
[4 marks]
Notes: Do not award any marks if there is clear evidence of adding instead of multiplying.
Do not award final A1 for a final answer of 108864, even if 108864x3 is seen previously.
If no working shown award N2 for 108864.
Examiners report
This is a common question and yet it was not unusual to see candidates writing out the expansion in full or using Pascal’s triangle to find the correct binomial coefficient. Of those candidates who managed to identify the correct term, many omitted the parentheses around 2χ which led to an incorrect answer. Most candidates were able to distinguish between “the term in x3n” and the coefficient. There are still a significant number of candidates who add the parts of a term rather than multiply them and this approach gained no marks.
This is a common question and yet it was not unusual to see candidates writing out the expansion in full or using Pascal’s triangle to find the correct binomial coefficient. Of those candidates who managed to identify the correct term, many omitted the parentheses around 2χ which led to an incorrect answer. Most candidates were able to distinguish between “the term in x3n” and the coefficient. There are still a significant number of candidates who add the parts of a term rather than multiply them and this approach gained no marks.