Sigma Notation
1.
Using a non-calculator method, find the set of values of 𝑘 for which ∑𝑘𝑟=3 (2𝑟 + 𝑘) > 𝑘 2.
[5]
[2015/TJC/MCT/2]
2.
2
3
1
𝑟+4
(i) Show that 𝑟 − 𝑟+1 + 𝑟+2 ≡ 𝑟(𝑟+1)(𝑟+2).
[2]
3𝑟+12
(ii) Hence find 𝑆𝑛 = ∑𝑛𝑟=1 𝑟(𝑟+1)(𝑟+2), giving your answer in the form 𝑘 − f(𝑛), where 𝑘 is a
constant.
9
(iii) Deduce that 𝑆𝑛 < 2, and give a reason why 𝑆∞ is convergent.
[2]
[2008/SAJC/Prelims/I(b)]
3.
The sum, 𝑆𝑛 , of the first 𝑛 terms of a sequence 𝑢1 , 𝑢2 , 𝑢3 , … is given by
𝑆𝑛 =
3
2𝑛 + 1
−
2 𝑛(𝑛 + 1)
(i) Give a reason why the series ∑𝑢𝑟 converges, and write down the sum to infinity.
(ii) Find a formula for 𝑢𝑛 .
(iii) Determine, with a reason, if the sequence {𝑢𝑛 } is a geometric progression.
[2]
[2]
[2]
4.
2
Express 𝑟(𝑟2 −1) in partial fractions.
1
1
[3]
1
1
Given that 𝑆𝑁 = (4)(15) + (5)(24) + (6)(35) + ⋯ + (𝑁)(𝑁2 −1), find 𝑆𝑁 in terms of 𝑁.
Deduce the limit of 𝑆𝑁 as 𝑁 → ∞.
[4]
[2008/RIJC/Prelims/I/2]
5.
The sum of the first 𝑛 terms of a progression is given by 𝑆𝑛 = 1 − e2𝑛 .
(i) Find 𝑈𝑛 , the 𝑛 th term of the progression.
[2]
(ii) Prove that 𝑈1 , 𝑈2 , 𝑈3 , … is a geometric progression. Hence determine if the sum
𝑆𝑛 converges.
[3]
[modified 2014/RVHS/promos/2]
6.
The sequence of real numbers 𝑏1 , 𝑏2 , 𝑏3 , … is such that 𝑏1 = 4 and 𝑏𝑛 = 𝑏𝑛−1 + 2𝑛 for
all 𝑛 ∈ ℤ+ , 𝑛 ≥ 2.
Using the method of difference, show that 𝑏𝑛 = 𝑛2 + 𝑛 + 2 for all 𝑛 ∈ ℤ+ .
[3]
[2010/ACJC/Prelim/II/3]
7.
The sum, 𝑆𝑛 , of the first 𝑛 terms of a sequence 𝑢1 , 𝑢2 , 𝑢3 , … is given by
𝑆𝑛 = 1 − 𝑛 − 2−𝑛
(i) Find 𝑢𝑛 .
[2]
(ii) Determine if the following series is a convergent series. For the series that converge,
write down the value of the sum to infinity.
(a) ∑∞
𝑛=1 𝑢𝑛
(b) ∑∞
[4]
𝑛=1 (𝑢𝑛 + 1)
8.
𝑟 2 +7𝑟+11
𝐴
𝐵
in the form
+
(𝑟+4)!
(𝑟+2)!
(𝑟+4)!
𝑟 2 +7𝑟+11
5
𝑛+5
show that ∑𝑛𝑟=1
= −
.
(𝑟+4)!
4!
(𝑛+4)!
2 +9𝑟+19
𝑟
(ii) Hence, find ∑∞
𝑟=1 (𝑟+5)! .
(i) By expressing
where 𝐴 and 𝐵 are real constants,
[3]
[3]
[2010/NJC/ Prelims /II/2( b)]
9.
(i) Given that 1 − 2𝑟 = 𝐴(𝑟 + 1) + 𝐵𝑟, find the constants 𝐴 and 𝐵.
1−2𝑟
(ii) Use the method of differences to find ∑𝑛𝑟=1 3𝑟 .
[1]
[3]
(iii) Hence find the value of ∑∞
𝑟=1
[4]
2−2𝑟
.
3𝑟
[2016/ CJC/Prelims/I/6]
10.
𝑟
)
𝑟+2
𝑟
(a) Using the method of difference or otherwise, evaluate ∑2𝑛
𝑟=1 (−1) ln (
𝑟
∑2𝑛−1
𝑟=1 (−1) ln
𝑟
( ),
𝑟+2
and
giving your answer as a single logarithmic function in 𝑛.
[4]
𝑟
𝑟
Explain why the series ∑∞
𝑟=1 (−1) ln (𝑟+2) is a convergent series, and state the sum to
infinity in the form ln 𝑎, where 𝑎 is a constant to be determined.
[2]
[2015/AJC/promos/11a]
(b) A sequence of numbers 𝑢1 , 𝑢2 , 𝑢3 , … has a sum 𝑆𝑛 where 𝑆𝑛 = ∑𝑛𝑟=1 𝑢𝑟 = 𝑎𝑛2 + 𝑏𝑛,
where 𝑎 and 𝑏 are constants. Given that the first term and tenth term of the sequence are
1 and 19 respectively, find the values of 𝑎 and 𝑏.
[3]
11.
A sequence 𝑢1 , 𝑢2 , 𝑢3 , … is such that 𝑢1 = 3 and 𝑢𝑛+1 = 𝑢𝑛 + 4𝑛 + 2 −
(i) Express
1
𝑟(𝑟+1)
1
,
𝑛(𝑛+1)
for all 𝑛 ∈ ℤ+
in partial fractions. Hence, by using the method of differences, find
1
∑𝑛−1
𝑟=1 (4𝑟 + 2 − 𝑟(𝑟+1)).
[4]
(ii) By considering 𝑢𝑟+1 − 𝑢𝑟 and using the result in part (i), show that
1
𝑢𝑛 = 2𝑛2 + , where 𝑛 ∈ ℤ+ .
𝑛
[3]
[2011/HCI/ Prelims /I/8]
12.
1
2𝑟−1
Given f(𝑟) = 𝑟2 , 𝑟 ≠ 1, show that f(𝑟 − 1) − f(𝑟) = 𝑟2 (𝑟−1)2 .
3
5
7
(i) Hence or otherwise, find the sum to infinity of the series 22 (1)2 + 32 (2)2 + 42 (3)2 + ⋯ [3]
2𝑟−1
(ii) Deduce that ∑𝑛𝑟=2 𝑟2 (𝑟+1)2 < 1.
[2]
[2009/SRJC/ Prelims /I/14(a)]
13.
1
1
𝑛+𝑎
1
(i) Given that 2(𝑛−1)2 − 2𝑛2 = 𝑛2 (𝑛−1)2, show that 𝑎 = − 2.
(ii) Given that 𝑆𝑁 =
2𝑛−1
∑𝑁
𝑛=𝑀 2𝑛2 (𝑛−1)2,
[1]
state the smallest possible value of 𝑀, where
+
𝑀 ∈ ℤ and 𝑀 ≤ 𝑁, such that 𝑆𝑁 can be defined.
(iii) If 𝑀 = 3, find 𝑆𝑁 in terms of 𝑁.
1
1
1
(iv) Deduce that the sum to infinity of the series (2)(32 ) + (3)(42 ) + (4)(52 ) + ⋯ is
1
less than 8.
[1]
[3]
[3]
[2010/RIJC/Prelims/II/1]
14.
Verify that
1
2𝑥
1
1
+
(𝑥−1)
2(𝑥−2)
1
.
𝑥(𝑥−1)(𝑥−2)
1
∑𝑁
𝑛=3 𝑛(𝑛−1)(𝑛−2).
=
[1]
By using the above result, find
[4]
−
Hence show that
1
∑∞
𝑛=1 𝑛3
<
11
.
8
[3]
15.
1
Let 𝑈𝑟 = (𝑟+1)(𝑟+2)(𝑟+3)⋯(𝑟+𝑝) where 𝑝 is a positive integer.
Simplify 𝑈𝑟 − 𝑈𝑟−1 as far as possible.
1
Hence prove that ∑𝑛𝑟=1
Deduce the exact
=
1 1
1
[ − (𝑛+1)(𝑛+2)⋯(𝑛+𝑝)]
𝑝 𝑝!
𝑟(𝑟+1)(𝑟+2)⋯(𝑟+𝑝)
1
value of ∑∞
𝑟=0 (𝑟+1)(𝑟+2)(𝑟+3)(𝑟+4).
⋅
[2]
[4]
[2]
[2008/NYJC/ Prelims /I/5]