Edexcel P2 2022 June — Question 1 4 marks

Exam BoardEdexcel
ModuleP2 (Pure Mathematics 2)
Year2022
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeExpansion up to x^3 term
DifficultyEasy -1.2 This is a straightforward application of the binomial theorem requiring only direct substitution into the formula (a+b)^n with positive integer n=10. Students need to calculate four terms using nCr and simplify fractions to integers, but no problem-solving or conceptual insight is required—purely mechanical execution of a standard algorithm.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n
Find the first four terms, in ascending powers of \(x\), of the binomial expansion of $$\left(2 + \frac{3}{8}x\right)^{10}$$ Give each coefficient as an integer. [4]
Question 1:
AnswerMarks Guidance
11 0 2 4 +  B1
 + 1 0 C   x + 1 0 C   x 2 + 1 0 C   x 3 + 
AnswerMarks
1 2 3M1
2 3
3 3  3 
+1029 x+4528  x +12027  x
8 8  8 
Or two of
AnswerMarks
+1920x+1620x2 +810x3A1
 + 1 9 2 0 x + 1 6 2 0 x 2 + 8 1 0 x 3A1
(4)
(4 marks)
Notes:
B1: Correct constant term of 1024 as an integer.
M1: Correct binomial coefficient multiplied by the correct powers of x for at least 2 terms in x, x2
or x3.
 1 0   1 0   1 0 
Allow e.g. 10C , 10C , 10C or , , or evaluated coefficients and condone missing
1 2 3
1 2 3
2 3
3  3  3  3 
brackets e.g. x 2 for x or x3 for x
8 8 8 8
May take out a common factor of 210 first, but again look for correct binomial coefficients
multiplied by the correct powers of x for at least 2 terms in x, x2 or x3.
1 0 1 0
E.g.  2 + 3 x  = 2 1 0  1 + 3 x  = 2 1 0 ( 1 + 1 0 C . . . x + 1 0 C . . . x 2 + 1 0 C . . . x 3 + . . . )
8 1 6 1 2 3
A1: Allow for a fully correct unsimplified expression (ignoring the constant term) with evaluated
binomial coefficients, must be expanded if a common factor of 210 is taken out first
OR two of the three terms in x, x2 and x3 correct and simplified.
2 3
3 3  3 
E.g. +1029 x+4528  x +12027  x
8 8  8 
2 3
3  3   3 
Or  + 1 0  2 1 0  x + 4 5  2 1 0  x + 1 2 0  2 1 0  x
1 6 1 6 1 6
The brackets must be present unless they are implied by subsequent work.
+1920x+1620x2 +810x3
OR two of . Allow terms to be “listed”.
A1: Final three terms fully correct and simplified. Allow terms to be “listed”.
Once a correct expansion (or list of terms) is seen then isw. E.g. some candidates think they
have to list the coefficients separately but apply isw.
Ignore any extra terms if found.
For reference incorrect bracketing:
. . . + 4 5  2 8 
3
8
x 2 + 1 2 0  2 7 
3
8
x 3 gives .. . + 4 3 2 0 x 2 + 5 7 6 0 x 3
PMT
And usually scores B1M1A0A0 if the 1024 is correct
Special case: Some candidates are just finding the coefficients 1024 + 1920 + 1620 + 810 and this scores B1
only if the x’s never make an appearance.
AnswerMarks Guidance
QuestionScheme Marks 
Question 1:
1 | 1 0 2 4 +  | B1
 + 1 0 C   x + 1 0 C   x 2 + 1 0 C   x 3 + 
1 2 3 | M1
2 3
3 3  3 
+1029 x+4528  x +12027  x
8 8  8 
Or two of
+1920x+1620x2 +810x3 | A1
 + 1 9 2 0 x + 1 6 2 0 x 2 + 8 1 0 x 3 | A1
(4)
(4 marks)
Notes:
B1: Correct constant term of 1024 as an integer.
M1: Correct binomial coefficient multiplied by the correct powers of x for at least 2 terms in x, x2
or x3.
 1 0   1 0   1 0 
Allow e.g. 10C , 10C , 10C or , , or evaluated coefficients and condone missing
1 2 3
1 2 3
2 3
3  3  3  3 
brackets e.g. x 2 for x or x3 for x
8 8 8 8
May take out a common factor of 210 first, but again look for correct binomial coefficients
multiplied by the correct powers of x for at least 2 terms in x, x2 or x3.
1 0 1 0
E.g.  2 + 3 x  = 2 1 0  1 + 3 x  = 2 1 0 ( 1 + 1 0 C . . . x + 1 0 C . . . x 2 + 1 0 C . . . x 3 + . . . )
8 1 6 1 2 3
A1: Allow for a fully correct unsimplified expression (ignoring the constant term) with evaluated
binomial coefficients, must be expanded if a common factor of 210 is taken out first
OR two of the three terms in x, x2 and x3 correct and simplified.
2 3
3 3  3 
E.g. +1029 x+4528  x +12027  x
8 8  8 
2 3
3  3   3 
Or  + 1 0  2 1 0  x + 4 5  2 1 0  x + 1 2 0  2 1 0  x
1 6 1 6 1 6
The brackets must be present unless they are implied by subsequent work.
+1920x+1620x2 +810x3
OR two of . Allow terms to be “listed”.
A1: Final three terms fully correct and simplified. Allow terms to be “listed”.
Once a correct expansion (or list of terms) is seen then isw. E.g. some candidates think they
have to list the coefficients separately but apply isw.
Ignore any extra terms if found.
For reference incorrect bracketing:
. . . + 4 5  2 8 
3
8
x 2 + 1 2 0  2 7 
3
8
x 3 gives .. . + 4 3 2 0 x 2 + 5 7 6 0 x 3
PMT
And usually scores B1M1A0A0 if the 1024 is correct
Special case: Some candidates are just finding the coefficients 1024 + 1920 + 1620 + 810 and this scores B1
only if the x’s never make an appearance.
Question | Scheme | Marks
Find the first four terms, in ascending powers of $x$, of the binomial expansion of

$$\left(2 + \frac{3}{8}x\right)^{10}$$

Give each coefficient as an integer.
[4]

\hfill \mbox{\textit{Edexcel P2 2022 Q1 [4]}}
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