Series Cheat Sheet Edexcel Core Pure 1

You need to be comfortable in expressing a series using sigma notation.

1
Inputting 𝑟 = 1 into this expression gives us the first Example 2: Show that ∑𝑛𝑟=1 𝑟(𝑟 + 3)(2𝑟 − 1) = 6 𝑛(𝑛 + 1)(3𝑛2 + 𝑎𝑛 + 𝑏), where 𝑎 and 𝑏 are integers to be found.
This tells us the last value of r for term, 𝑟 = 2 gives us the second term, and so forth.
our sequence, i.e. our last term will Expanding the brackets: 𝑛 𝑛
4 ∑ 𝑟(𝑟 + 3)(2𝑟 − 1) = ∑ 2𝑟 3 + 5𝑟 2 − 3𝑟
be 7 − 2(4) = −1
∑(7 − 2𝑟) = 5 + 3 + 1 − 1 = 8 𝑟=1 𝑟=1
This series in particular is an
This is the value of r where our 𝑟=1 Manipulating the sum: 𝑛 𝑛 𝑛
arithmetic series which is introduced
series starts, i.e. our first term is
in Pure Year 2. = 2 ∑ 𝑟3 + 5 ∑ 𝑟2 − 3 ∑ 𝑟
7 − 2(1) = 5. 𝑟=1 𝑟=1 𝑟=1

Using the results for ∑ 𝑟 , ∑ 𝑟 2 , ∑ 𝑟 3 : 1 1 𝑛

= 2 ൤ 𝑛2 (𝑛 + 1)2 ൨ + 5 ൤ 𝑛(𝑛 + 1)(2𝑛 + 1)൨ − 3 ቂ (𝑛 + 1)ቃ
You can use the following rules to manipulate expressions involving sigma notation: 4 6 2
1
𝑛 𝑛 Simplifying before factoring out 6 𝑛(n+1): 1 2 5 3𝑛
= 𝑛 (𝑛 + 1)2 + 𝑛(𝑛 + 1)(2𝑛 + 1) − (𝑛 + 1)
∑𝑛 𝑘𝑓(𝑟) = 𝑘 ∑ 𝑓(𝑟) 2 6 2
▪ 𝑟=1 𝑘𝑓(𝑟) = 𝑘
∑
𝑟=1 𝑟=1 1
= 𝑛(𝑛 + 1)[3𝑛(𝑛 + 1) + 5(2𝑛 + 1) − 9]
𝑛 𝑛 𝑛 6

▪ . 𝑓(𝑟) + 𝑔(𝑟) = ∑ 𝑓(𝑟) + ∑ 𝑔(𝑟)

∑ Simplifying: 1
= 𝑛(𝑛 + 1)[3𝑛2 + 13𝑛 − 4]
𝑟=1 𝑟=1 𝑟=1 6

You can use the following results to evaluate some complicated series:
𝑛 1
You will not be given these. You will only be given Example 3: (a) Show that ∑𝑛𝑟=1(3𝑟 − 2)2 = 2 𝑛(6𝑛2 − 3𝑛 − 1).
𝑛1 = 𝑛
▪ ∑
∑𝑟=1 1 = 𝑛 results for ∑𝑛𝑟=1 𝑟 3 and ∑𝑛𝑟=1 𝑟 3 . 𝑟𝜋
𝑟=1 (b) Hence find any values of 𝑛 for which ∑𝑛𝑟=5(3𝑟 − 2)2 + 103 ∑20 3
𝑟=1 𝑟𝑐𝑜𝑠 ቀ 2 ቁ = 3𝑛 .

𝑛
1 1 (a) Expanding the brackets: 𝑛 𝑛
𝑛 𝑟 = 𝑛(𝑛 + 1)
▪ ∑
∑𝑟=1 𝑟 2= 𝑛 ∑(3𝑟 − 2)2 = ∑ 9𝑟 2 − 12𝑟 + 4
𝑟=1
2
𝑟=1 𝑟=1

𝑛 Manipulating the sum: 𝑛 𝑛 𝑛

1 2 = 9 ∑ 𝑟 2 − 12 ∑ 𝑟 + 4 ∑ 1
▪ .∑ 𝑟 = 6 𝑛(𝑛 + 1)(2𝑛 + 1) 𝑟=1 𝑟=1 𝑟=1
𝑟=1
You can prove these results using the proof by induction method
𝑛 covered in Chapter 8 of Core Pure 1. Using the results for ∑ 1 , ∑ 𝑟 , ∑ 𝑟 2 : 9 𝑛
1 21 2 2 = 𝑛(𝑛 + 1)(2𝑛 + 1) − 12 ቂ (𝑛 + 1)ቃ + 4(𝑛)
6 2
▪ ∑𝑛𝑟=1
∑ 𝑟 34=
𝑟3 = 𝑛 (𝑛𝑛+ 1)
4
𝑟=1 Simplifying: 3𝑛
= (2𝑛2 + 3𝑛 + 1) − 6𝑛(𝑛 + 1) + 4𝑛
2
To find the sum of a series that does not start at 𝑟 = 1, you can instead use the following result: 𝑛
Factoring out 2 : 𝑛 𝑛
= [6𝑛2 + 9𝑛 + 3 − 12𝑛 − 12 + 8) = [6𝑛2 − 3𝑛 − 1]
𝑛 𝑛 𝑘−1
2 2
𝑛 𝑛 𝑘−1
▪ ∑𝑟=𝑘
∑ 𝑓(𝑟)𝑓(𝑟) ∑𝑟=1
= ∑=𝑓(𝑟) −∑ 𝑓(𝑟)− ∑𝑟=1 𝑓(𝑟)
𝑓(𝑟) Dealing with the first term to begin with; the sum starts at 𝑟 = 5 so we 𝑛 𝑛 4

𝑟=𝑘 𝑟=1 𝑟 =1 need to use the result ∑𝑛𝑟=𝑘 𝑓(𝑟) = ∑𝑛𝑟=1 𝑓(𝑟) − ∑𝑘−1
𝑟=1 𝑓(𝑟) ∑(3𝑟 − 2)2 = ∑(3𝑟 − 2)2 − ∑(3𝑟 − 2)2
𝑟=5 𝑟=1 𝑟=1
𝑛
Using the result from part a: 𝑛 4
∑(3𝑟 − 2) = [6𝑛2 − 3𝑛 − 1] − [6(4)2 − 3(4) − 1]
2
1 2 2
Example 1: (a) Show that ∑𝑛𝑟=1(7𝑟 − 4) = 2 𝑛(7𝑛 − 1). 𝑟=5

(b) Hence evaluate ∑50 3𝑛2 𝑛

𝑟=20(7𝑟 − 4). = 3𝑛3 − − − 166
2 2
𝑛 𝑛 𝑛
(a) Manipulating the sum: Dealing with the second term: ∑20
𝑟𝜋 𝑟𝜋
𝑟=1 𝑟𝑐𝑜𝑠 ቀ 2 ቁ has a periodic nature since 𝑐𝑜𝑠 ቀ 2 ቁ will be zero for odd 𝑟 and
∑(7𝑟 − 4) = 7 ∑ 𝑟 − 4 ∑ 1
for even 𝑟 it will either be 1 or −1. Simply by writing out the first few terms,
𝑟=1 𝑟=1 𝑟=1 You can manually calculate the sum once you identify the periodic nature of
we can see what this sum will be:
1 the sum.
Using the result ∑𝑛𝑟=1 𝑟 = 2 𝑛(𝑛 + 1). 𝑛 1
= 7 ቂ (𝑛 + 1)ቃ − 4𝑛 = 𝑛[7(𝑛 + 1) − 8)] 20
2 2 𝑟𝜋
∑ 𝑟𝑐𝑜𝑠 ቀ ቁ = 0 − 1 + 0 + 3 + 0 − 5 + ⋯ − 18 + 0 + 20
Simplifying: 1 2
= 𝑛[7𝑛 − 1] 𝑟=1
2 Adding up the terms: Adding up the terms manually gives ∑20
𝑟𝜋
𝑟=1 𝑟𝑐𝑜𝑠 ቀ 2 ቁ = 14.
(b) Using the above result to find the sum of a series that does not start at 50 50 19

∑ (7𝑟 − 4) = ∑(7𝑟 − 4) − ∑(7𝑟 − 4) Substituting this result back into the given equation: 3𝑛2 𝑛
𝑟 = 1: 3𝑛3 − − − 166 + 103(14) = 𝑛3
𝑟=20 𝑟=1 𝑟=1 2 2
Simplifying gives us a quadratic: 3𝑛2 𝑛
+ − 1276 = 0
Using our part (a) result with 𝑛 = 50 for the first sum and 𝑛 = 19 for the 1 1 2 2
second sum: = (50)[7(50) − 1] − (19)[7(19) − 1] Solving the quadratic using the quadratic formula: 88
2 2 Quadratic formula: 𝑛 = 29 or 𝑛 = − 3 . Term number must be a positive
= 7471 You could also factorise this quadratic. integer so 𝑛 = 29.

https://bit.ly/pmt-cc
https://bit.ly/pmt-edu https://bit.ly/pmt-cc