C1 Sequences and Series - General
C1 Sequences and Series - General
C1 Sequences and Series - General
com
Paper Reference(s)
6663/01
Edexcel GCE
Core Mathematics C1
Advanced Subsidiary
A booklet ‘Mathematical Formulae and Statistical Tables’ might be needed for some questions.
The marks for the parts of questions are shown in round brackets, e.g. (2).
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.
1
C1 Sequences and series – General PhysicsAndMathsTutor.com
a n +1 = (a n2 + 3 , n ≥ 1,
a1 = 2
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2. A sequence a 1 , a 2 , a 3 , … is defined by
a 1 = k,
a n+1 = 2a n – 7, n ≥ 1,
where k is a constant.
4
Given that ∑a
r =1
r = 43,
___________________________________________________________________________
3. A sequence x 1 , x 2 , x 3 , … is defined by
x 1 = 1,
x n + 1 = ax n – 3, n > 1,
where a is a constant.
Given that x 3 = 7,
___________________________________________________________________________
x 1 = 1,
Given that x 3 = 1,
___________________________________________________________________________
a 1 = k,
a n+1 = 3an + 5, n ≥ 1,
4
(c) (i) Find ∑a
r =1
r in terms of k.
4
(ii) Show that ∑a
r =1
r is divisible by 10.
(4)
(Total 7 marks)
___________________________________________________________________________
6. A sequence a 1 , a 2 , a 3 , … is defined by
a1 = 3
a n + 1 = 3a n – 5, n ≥ 1.
5
(b) Calculate the value of ∑a
r =1
r
(3)
(Total 5 marks)
___________________________________________________________________________
Given that u 3 = 3u 2 ,
___________________________________________________________________________
un a
un + 1 = + , n = 1, 2, 3, ...,
2 un
where a is a constant.
(a) Given that a = 20 and u 1 = 3, find the values of u 2 , u 3 and u 4 , giving your answers
to 2 decimal places.
(3)
___________________________________________________________________________
a1 = k , a n +1 = 4 a n − 7 ,
where k is a constant.
___________________________________________________________________________
1. (a)
a2 = ( 4+3 =) 7
B1
= a3 "their=
7"+ 3 10 B1ft 2
Note
2nd B1ft follow through their “7” in correct formula provided they
have n , where n is an integer.
(b)
a4 = (
10 + 3 = 13 ) M1
a5= 13 + 3= 4 * A1 cso 2
Note
M1 for an attempt to find a4. Should see " their" (a3 ) 2 + 3 .
Must see evidence for M1.
a4 = 13 provided this follows from their a3 working or answer is sufficient
A1cso for a correct solution (M1 explicit) must include the = 4.
Ending at 16 only is A0 and ending with ± 4 is A0.
Ignore any incorrect statements that are not used e.g. common
difference = 3
Listing: A full list: 2 (= 4 ), 7 , 10 , 13 , 16 = 4 is fine for M1A1
ALT
Formula: Some may state (or use) an = 3n + 1 leading to a5 = 3 × 5 + 1 = 4.
This will get marks in (a) [if correct values are seen] and can
score the M1 in (b) if an = 3n + 1 or a4 = 13 are seen.
∑a
r =1
r = k + “(2k – 7)” + (4k – 21) + “(8k – 49)” M1
1st M1 for an attempt to find a4 using the given rule. Can be awarded
for 8k – 49 seen.
Use of formulae for the sum of an arithmetic series scores M0M0A0 for
the next 3 marks.
2nd M1 for attempting the sum of the 1st 4 terms. Must have “+” not
just, or clear attempt to sum.
Follow through their a2 and a4 provided they are linear
functions of k.
Must lead to linear expression in k. Condone use of their linear
a3 ≠ 4k – 21 here too.
3rd M1 for forming a linear equation in k using their sum and the 43 and
attempt to solve for k as far as pk = q
120
A1 for k = 8 only so k = is A0
15
Answer Only (e.g. trial improvement)
Accept k = 8 only if 8 + 9 + 11 + 15 = 43 is seen as well
Sum a 2 + a 3 + a 4 + a 5 or a 2 + a 3 + a 4 +
Allow: M1 if 8k – 49 is seen, M0 for the sum (since they are not adding
the 1st 4 terms) then M1
if they use their sum along with the 43 to form a linear equation and
attempt to solve but A0
[7]
3. (a) [x2 =] a – 3 B1 1
B1 for a × 1 – 3 or better. Give for a – 3 in part (a) or if it
appears in (b) they must state x2 = a – 3
This must be seen in (a) or before the a(a – 3) – 3 step.
M1 for clear show that. Usually for a(a – 3) – 3 but can follow
through their x2 and even allow ax2 – 3
A1 for correct processing leading to printed answer.
Both lines needed and no incorrect working seen.
(c) a2 – 3a – 3 = 7
a2 – 3a – 10 = 0 or a2 – 3a = 10 M1
(a – 5)(a + 2) = 0 dM1
a = 5 or – 2 A1 3
4. (a) 1(p + 1) or p +1 B1 1
(c) 1+ 3p + 2p2 =1 M1
p(2p + 3) = 0 p = ... M1
3
p= − (ignore p = 0, if seen, even if ‘chosen’ as the answer) A1 3
2
∑a
r =1
r = k + (3k + 5) + (9k + 20) + (27k + 65) M1
(ii) = 40k + 90 A1
= 10(4k + 9) (or explain why divisible by 10) A1ft 4
2nd M1 for attempting sum of 4 relevant terms, follow through their (a)
and (b).
Must have 4 terms starting with k.
Use of arithmetic series formulae at this point is M0A0A0
[7]
6. (a) a2 = 4 B1
a3 = 3 × a2 − 5 = 7 B1f.t. 2
2nd B1f.t. Follow through their a2 but it must be a value.
3× 4− 5 is B0
Give wherever it is first seen.
7. (a) u2 = (–1)(2) + d = –2 + d B1
u3 = (–1)2(–2 + d) + d = –2 + 2d M1
Attempting to find u3 in terms of d
u4 = (–1)3(–2 + 2d) + d = 2 – d A1 4
u3 and u4 correct
3 a 3 a
(b) (i) 3 = + or 9 = + M1
2 3 2 3
a 3 3
= 9 − or a = 3 9 − M1
3 2 2
a = 22.5 A1 3
M1 A correct equation for a, with or without .
M1 Attempt correct manipulation to ka = ,(k > 0).
9. (a) 4k − 7 B1
(b) 4 (4k − 7) − 7 = 16k − 35 M1 A1 2
(c) 16k − 35 = 13 k=3 M1 A1 2
[5]
1. This proved to be a straightforward question for most candidates who worked through it
carefully and gained full marks. A few noticed that the numbers inside the square root
formed an arithmetic sequence and this sometimes distracted them as they tried to use
formulae for arithmetic series.
There were still some candidates who did not understand the notation and interpreted
a n as n 2 + 3 .
Some were confused by the nested square roots and we saw a 3 = 7 2 + 3 and others
thought ( 7)2
= 49 but overall this question was answered well.
2. There were far fewer cases of candidates not understanding how an inductive formula
like this works and many were able to answer parts (a) and (b) successfully. Part (b)
required the candidates to “show” a given result and most gave the expression 2(2k – 7)
– 7 which was fine but a small minority thought the pattern must be 2 × 2k – 3 × 7. Part
(c) met with mixed success: many found a4 but some then solved a4 = 43 whilst others
assumed that the series was arithmetic and attempted to use a formula such as
4
(a1 + z 4 ). Those who did attempt the correct sum occasionally floundered with the
2
arithmetic but there were plenty of fully correct solutions seen.
3. The notation associated with sequences given in this form still causes difficulties for
some candidates and as a result parts (a) and (b) were often answered less well than part
(c). A common error in the first two parts was to leave an x in the expression but most
of those who could handle the notation gave clear and accurate answers. There were the
usual errors in part (c), with a2 – 3a – 4 = 0 appearing quite often and it was
encouraging to see most candidates factorising their quadratic expression confidently as
a means of solving the equation. A few candidates still use a trial and improvement
approach to questions of this type and they often stopped after finding just one solution
and gained no credit.
4. Although just a few candidates failed to understand the idea of the recurrence relation,
most managed to complete the first two parts successfully. A major concern in part (b),
however, was the widespread lack of brackets in the algebraic expressions. It was
usually possible for examiners to interpret candidates’ intentions generously, but there
needs to be a greater awareness that, for example, 1 + p(1 + 2p) is not an acceptable
alternative to (1 + p)(1 + 2p).
The given answer to part (b) enabled the vast majority of candidates to start part (c)
correctly, but the main problem with this part was in solving 2p 2 + 3p = 0, which
proved surprisingly difficult for some. Attempts to complete the square usually failed,
while the quadratic formula method, although generally more successful, often suffered
from mistakes related to the fact that c was zero. Those who did manage to factorise the
3 3
expression sometimes gave the answer p= instead of p = − . It was clear that
2 2
candidates would have been much happier solving a 3-term quadratic equation. Those
who trivialised the question by giving only the zero solution (despite the condition p >
0) scored no further marks in the question.
Part (d) proved challenging for many candidates. Some used the solution p = 0 and
some tried to make use of the sum formula for an arithmetic series. Few candidates were
successful, but those who wrote out the first few terms were more likely to spot the
‘oscillatory’ nature of the sequence. Good candidates stated that even terms were all
1 1
equal to − and therefore the 2008th term was − . Quite a large number of candidates
2 2
were able to express x2008 in terms of x2007, but those who simply substituted 2007
into one of their expressions often wasted time in tedious arithmetic that led to a very
large answer.
5. Many of the comments made on the June 2006 paper would apply here too. Many
candidates were clearly not familiar with the notation and a number used arithmetic
series formulae to find the sum in part (c) although this was less common than in June
2006.
Apart from those candidates who had little idea about this topic most were able to
answer parts (a) and (b) correctly. In part (c) many attempted to find a4 using the
recurrence relation and those who were not tempted into using the arithmetic series
formulae often went on to attempt the sum and usually obtained 40k + 90 which they
were easily able to show was divisible by 10.
Some lost marks for poor arithmetic 30k + 90 and 40k + 80 being some of the incorrect
answers seen.
6. Most candidates knew how to start this question and full marks for part (a) was
common, however some lost out due to poor arithmetic such as 3 × 3 – 5 = 6 – 5 = 1 for
a2 A minority of candidates though had no idea how to interpret the recurrence relation
notation with a significant number interpreting 3a –5 as 3 × n – 5. In part (b) many
candidates were convinced that the series had to be arithmetic and they gained no
further marks. Some did find a4 + a5 and a correctly but then used the arithmetic
n( a + l )
formula with l= 43 Clearly students are familiar with the work on arithmetic
2
series but in some cases this seems to have overshadowed their understanding of
recurrence relations.
8. This question was answered very well and many scored full marks. In part (a) some
failed to give answers to 2 decimal places and sometimes the wrong substitution was
made or the square root omitted. Part (b)(ii) caused a few problems, some candidates
did not appreciate the instruction to “write down” and worked through u3, and u4
before stating the answer. Others started with 2.89 instead of 3.