What is 4.3: Sigma notation?

The notation _r=a^bf(r) means add the values of f(r) for each integer r from a to b inclusive.

How many marks are 4.3 questions worth?

4.3 practice drills on this page are worth 2–5 marks each; marks in a real exam vary by question.

Which past exam papers has 4.3 appeared in?

We have not yet indexed a verified past-paper appearance for 4.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

4.3

Understand and use sigma notation for sums of series.

Draft — not yet indexed

Sigma notation

Worked answers and methods for 4.3 on Edexcel A-level Maths 9MA0.

Explanation

The notation $\sum_{r=a}^{b}f(r)$ means add the values of $f(r)$ for each integer $r$ from $a$ to $b$ inclusive.
To write a series in sigma notation, identify a formula for its general term together with the correct first and last index values.
Use linearity to split sums: for example, $\sum_{r=1}^{n}(ar+b)=a\sum_{r=1}^{n}r+b\sum_{r=1}^{n}1$ , with $\sum_{r=1}^{n}1=n$ .
Check the endpoint terms after changing an index.
A common error is to treat the upper limit as the number of terms when the lower limit is not $1$ .

Worked example

Write the series $5+9+13+\cdots+41$ using sigma notation.

1.The terms have common difference $4$ , and $4r+1$ gives $5$ when $r=1$ .
2.Solving $4r+1=41$ gives $r=10$ , so the series is $\sum_{r=1}^{10}(4r+1)$ .

Answer: $\sum_{r=1}^{10}(4r+1)$

Common mistakes

Don't shift the index but leave the summand unchanged, so the re-indexed sum represents different terms.
Don't write a sigma expression whose lower and upper limits produce the wrong first or last term.

Exam tip

Check a sigma answer by substituting both endpoint indices and confirming the number of generated terms.

Worked practice

Tier 1 · Easy

Evaluate

\sum_{r=1}^{4}(2r+1)

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$24$	2
Notes
Substitute $r=1,2,3,4$ : the sum is $3+5+7+9=24$ .

(2 marks)

Tier 2 · Standard

Evaluate

\displaystyle\sum_{r=1}^{8}(3r^2-2r)

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$540$	3
Notes
Evaluate the eight terms directly. For $r=1,2,\ldots,8$ , the values of $3r^2-2r$ are $1,8,21,40,65,96,133,176$ . Their sum is $1+8+21+40+65+96+133+176=540$ .

(3 marks)

Tier 3 · Hard

Given that

\sum_{r=1}^{n}(4r-1)=210

, find the positive integer

n

. Show all stages of your working.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$n=10$	5
Notes
By linearity, $\sum_{r=1}^{n}(4r-1)=4[n(n+1)/2]-n=2n^2+n$ . Hence $2n^2+n=210$ , so $2n^2+n-210=0$ . Factorising gives $(2n+21)(n-10)=0$ . Since $n$ is positive, $n=10$ .

(5 marks)

Tier 1 · Easy

Write the series

11+16+21+\cdots+71

in the form

\displaystyle\sum_{r=1}^{n}(ar+b)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\displaystyle\sum_{r=1}^{13}(5r+6)$	2
Notes
The term $5r+6$ gives $11$ when $r=1$ . Solving $5r+6=71$ gives $r=13$ , so the series is $\sum_{r=1}^{13}(5r+6)$ .

(2 marks)

Tier 2 · Standard

Rewrite

\displaystyle\sum_{r=4}^{12}(7-2r)

as a sum whose lower limit is

1

. Hence evaluate the sum.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$\displaystyle\sum_{s=1}^{9}(1-2s)$ $-81$	4
Notes
Let $s=r-3$ . Then $r=4$ gives $s=1$ , $r=12$ gives $s=9$ , and $7-2r=7-2(s+3)=1-2s$ . Hence the sum is $\sum_{s=1}^{9}(1-2s)=9-2(9\times10/2)=-81$ .

(4 marks)

Tier 3 · Hard

Given that

\displaystyle\sum_{r=k}^{2k}(3r-1)=85

, where

k

is a positive integer, find

k

. Show all stages of your working.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$k=4$	5
Notes
There are $k+1$ terms, and $\sum_{r=k}^{2k}r=(k+2k)(k+1)/2=3k(k+1)/2$ . Therefore the given sum is $9k(k+1)/2-(k+1)=(k+1)(9k-2)/2$ . Equating this to $85$ gives $9k^2+7k-172=0=(k-4)(9k+43)$ . Since $k$ is a positive integer, $k=4$ .

(5 marks)

Tier 2 · Standard

Given that

\displaystyle\sum_{r=1}^{7}a_r=23

and

\displaystyle\sum_{r=1}^{7}b_r=-4

, evaluate

\displaystyle\sum_{r=1}^{7}(2a_r-3b_r+1)

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$65$	3
Notes
By linearity, the required sum is $2\sum_{r=1}^{7}a_r-3\sum_{r=1}^{7}b_r+\sum_{r=1}^{7}1$ . This is $2(23)-3(-4)+7=65$ .

(3 marks)

Tier 3 · Hard

Constants

a

and

b

satisfy the identity

\displaystyle\sum_{r=1}^{n}(ar+b)=n(2n+5)

for every positive integer

n

. Find

a

and

b

, showing how the identity determines both constants.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$a=4$ $b=3$	4
Notes
By linearity, the left side is $a\,n(n+1)/2+bn=(a/2)n^2+(a/2+b)n$ . The right side is $2n^2+5n$ . Since the identity holds for every positive integer $n$ , corresponding coefficients are equal. Thus $a/2=2$ , so $a=4$ , and then $a/2+b=5$ gives $b=3$ .

(4 marks)

Tier 3 · Hard

For a positive integer

n

, let

S_n=\displaystyle\sum_{r=1}^{2n}(-1)^r(3r-1)

. By grouping consecutive pairs of terms, show that

S_n=3n

. Obtain a corresponding expression for

\displaystyle\sum_{r=1}^{2n+1}(-1)^r(3r-1)

. Hence find

n

when this second sum is

-44

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$S_n=3n$ Odd-upper-limit sum $=-3n-2$ $n=14$	5
Notes
For each $j=1,\ldots,n$ , the terms with indices $2j-1$ and $2j$ sum to $-(6j-4)+(6j-1)=3$ . Hence $S_n=3n$ . The next term, with index $2n+1$ , is $-[3(2n+1)-1]=-(6n+2)$ , so the sum through $2n+1$ is $3n-(6n+2)=-3n-2$ . Setting this equal to $-44$ gives $-3n-2=-44$ , and therefore $n=14$ .

(5 marks)

Q10

Tier 3 · Hard

10.

The positive integer

n

satisfies

\displaystyle\sum_{r=1}^{n}|2r-7|=58

. Find

n

, showing how the modulus affects the sum.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$n=10$	5
Notes
For $r=1,2,3$ , the terms are $5,3,1$ , with sum $9$ . None of $n=1,2,3$ gives $58$ , so take $n\geq4$ . Then $\|2r-7\|=2r-7$ for $r\geq4$ , and $\sum_{r=4}^{n}(2r-7)=n(n+1)-12-7(n-3)=(n-3)^2$ . The equation is therefore $9+(n-3)^2=58$ , so $(n-3)^2=49$ . Since $n\geq4$ , $n-3=7$ and $n=10$ .

(5 marks)

Verified exam appearances

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