AMC/AIME/USA(J)MO Handout
Sequences and Series in the AMC and AIME
Author:
freeman66
nikenissan
For:
AoPS
Date:
February 1, 2021
n
n
1
A proof without words of the sum of the first n consecutive integers.
“Everything comes to you in perfect time, space and sequence.” – Louise Hay
freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
Contents
0 Acknowledgements 3
1 Introduction 4
2 Basics of Sequences and Series 4
2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Summation Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3 Arithmetic Sequences and Series 5
4 Geometric Sequences and Series 8
5 Arithmetico-Geometric Sequence 11
6 Telescoping Series 12
7 Recursive Sequences 14
8 Problems 15
A Appendix A: List of Theorems and Definitions 18
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
§0 Acknowledgements
This was made for the Art of Problem Solving Community out there! I would like to thank Evan Chen for his
evan.sty code. In addition, all problems in the handout were likely from the AoPS Wiki.
Art of Problem Solving Community
Evan Chen’s Personal Sty File
freeman66’s Website
nikenissan’s Website
And Evan says he would like this here for evan.sty:
Boost Software License - Version 1.0 - August 17th, 2003
Copyright (c) 2020 Evan Chen [evan at evanchen.cc]
https://web.evanchen.cc/ || github.com/vEnhance
He also helped with the hint formatting. Evan is a L
A
T
E
Xgod!
And finally, please do not make any copies of this document without referencing this original one. At least cite
us when you are using this document.
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
§1 Introduction
Sequences and series are one of the most prevalent topics in the AMC and AIME.
§2 Basics of Sequences and Series
§2.1 Definitions
Let’s start by introducing a few ideas.
Definition 2.1 (Sequence) — A sequence is an ordered list of numbers.
Definition 2.2
(Partial Sum)
—
A
partial sum
is the sum of a portion of the sequence. Partial sums are
sometimes called finite series.
Definition 2.3 (Series) — A series is the sum of the elements in a given sequence.
§2.2 Summation Notation
To keep things simple we introduce summation notation.
Definition 2.4
(Summation Notation)
—
In a summation, there is an index going from a starting point to
an ending point. The given element is tested through all values from the start to the end (inclusive), and
then all values of the element are summed.
Summation notation
is also sometimes called sigma notation.
An example would be
b
X
a=1
f(a) = f(1) + f(2) + ··· + f(b − 1) + f(b).
Example 2.5
Write 1 + 2 + . . . + n in summation notation.
Solution.
Denote
x
as the index of the summation. Notice the starting point of the summation is 1 and the
ending point of the summation is n. Our element is x in this scenario, thus our summation can be written as
n
X
x=1
x .
Theorem 2.6 (Distributive Property of Summations)
For all real constants k,
X
k · f(x) = k
X
f(x).
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Example 2.7
Evaluate
5
X
x=1
3 · x!
(x − 1)!
.
Solution. By the Distributive Property of Summations,
5
X
x=1
3 · x!
(x − 1)!
= 3
5
X
x=1
x!
(x − 1)!
.
Notice
x!
(x−1)!
= x. Thus,
3
5
X
x=1
x!
(x − 1)!
= 3
5
X
n=1
x.
P
5
x=1
x = 1 + 2 + 3 + 4 + 5 = 15. Therefore,
3
5
X
x=1
x = 3 · 15 = 45 .
Theorem 2.8 (Commutative Property of Summations)
For functions f (x) and g(x),
X
(f(x) ± g(x)) =
X
f(x) ±
X
g(x).
Summations simplify an expression a lot, and make it easier and more convenient to work with.
Exercise 2.9. Simplify
X
(x + 1)
2
−
X
x
2
.
Exercise 2.10. Compute the sum
∞
X
k=1
k
2
2
k
.
§3 Arithmetic Sequences and Series
Definition 3.1
(Arithmetic Sequence)
—
An
arithmetic sequence
is a sequence of numbers in which each
term is given by adding a fixed value to the previous term.
For example,
−
2
,
1
,
4
,
7
,
10
, . . .
is an arithmetic sequence because each term is three more than the previous
term. In this case, 3 is called the
common difference
of the sequence. More formally, an arithmetic sequence
a
n
is defined recursively by a first term
a
1
and
a
n
=
a
n−1
+
d
for
n ≥
2, where
d
is the common difference.
Explicitly:
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
Theorem 3.2 (Terms of an Arithmetic Sequence)
The n
th
term in an arithmetic sequence is described
a
n
= a
1
+ d(n − 1),
where a
n
is the n
th
term, a
1
is the first term, and d is the difference between consecutive terms.
Theorem 3.3 (Sum of an Arithmetic Sequence)
The sum of the first n terms of an arithmetic sequences is
s
n
=
n
2
(a
1
+ a
n
) =
n
2
(2a
1
+ (n − 1)d).
Proof. We can prove this with induction, which will not be taught or explained in this handout.
The following formulas are a direct consequence of the aforementioned theorems.
Corollary 3.4 (Sum of First n Positive Integers)
For all positive integers n,
1 + 2 + 3 + . . . + n =
n(n + 1)
2
.
Corollary 3.5 (Sum of First n Even Integers)
For all positive integers n,
2 + 4 + . . . + 2n = n(n + 1).
Corollary 3.6 (Sum of First n Odd Integers)
For all positive integers n,
1 + 3 + 5 + . . . + (2n − 1) = n
2
.
Example 3.7 (AIME I 2005/2)
For each positive integer
k,
let
S
k
denote the increasing arithmetic sequence of integers whose first term is 1
and whose common difference is
k.
For example,
S
3
is the sequence 1
,
4
,
7
,
10
, . . . .
For how many values of
k
does S
k
contain the term 2005?
Solution.
Suppose that the
n
th term of the sequence
S
k
is 2005. Then 1 + (
n −
1)
k
= 2005 =
⇒ k
(
n −
1) =
2004 = 2
2
·
3
·
167. Because
k
when multiplied by an integer is equal to 2004
,
in order words we can say that
k
is
a divisor/divides 2004
.
By the formula for the number of divisors, there are (2 + 1)(1 + 1)(1 + 1) = 12 divisors of
2
2
· 3
1
· 167
1
.
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
Example 3.8 (AIME II 2011/3)
The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence
with integer values. Find the degree measure of the smallest angle.
Solution.
The average angle in an 18-gon is 160
◦
. In this arrangement, the average is equivalent to the middle,
so the center two terms of the succession average to 160
◦
. In this manner for some certain (the grouping is
expanding and hence non-consistent) whole number
d
, the center two terms are (160
−d
)
◦
and (160 +
d
)
◦
. Since
the progression is 2
d
, the last term of the arrangement is (160 + 17
d
)
◦
, which is under 180
◦
, since the polygon
is raised. This gives 17
d <
20, so the lone appropriate positive whole number
d
is 1. The initial term is then
(160 − 17)
◦
= 143 .
Example 3.9 (AIME I 2012/2)
The terms of an arithmetic sequence add to 715. The first term of the sequence is increased by 1, the second
term is increased by 3, the third term is increased by 5, and in general, the
k
th term is increased by the
k
th
odd positive integer. The terms of the new sequence add to 836. Find the sum of the first, last, and middle
terms of the original sequence.
Solution.
If the sum of the original sequence is
n
X
i=1
a
i
then the sum of the new sequence can be expressed as
n
X
i=1
a
i
+ (2
i −
1) =
n
2
+
n
X
i=1
a
i
.
Therefore, 836 =
n
2
+ 715 =
⇒ n
= 11
.
Now the middle term of the original
sequence is simply the average of all the terms, or
715
11
= 65
,
and the first and last terms average to this middle
term, so the desired sum is simply three times the middle term, or 195 .
Example 3.10 (AMC 8 2015/18)
An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to
the previous term. Each row and each column in this 5
×
5 array is an arithmetic sequence with five terms.
What is the value of X?
X
17 81
1 25
(A) 21 (B) 31 (C) 36 (D) 40 (E) 42
Solution.
The middle term of the first row is
25+1
2
= 13 since the middle number is just the average in an
arithmetic sequence. Similarly, the middle of the bottom row is
17+81
2
= 49. Applying this again for the middle
column, the answer is
49+13
2
= 31 .
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
Exercise 3.11
(AHSME 1969/33)
.
Let
S
n
and
T
n
be the respective sums of the first
n
terms of two
arithmetic series. If
S
n
:
T
n
= (7
n
+ 1) : (4
n
+ 27) for all
n
, the ratio of the eleventh term of the first series
to the eleventh term of the second series is:
(A) 4 : 3 (B) 3 : 2 (C) 7 : 4 (D) 78 : 71 (E) undetermined
Exercise 3.12
(AIME 1999/1)
.
Find the smallest prime that is the fifth term of an increasing arithmetic
sequence, all four preceding terms also being prime.
§4 Geometric Sequences and Series
Definition 4.1
(Geometric Sequence)
—
A
geometric sequence
is a sequence of numbers in which each
term is a fixed multiple of the previous term.
For example: 1
,
2
,
4
,
8
,
16
,
32
, . . .
is a geometric sequence because each term is twice the previous term. In
this case, 2 is called the common ratio of the sequence. More formally, a geometric sequence may be defined
recursively by:
a
n
= r · a
n−1
, n > 1,
with a fixed first term a
1
and common ratio r. Using this definition, the nth term has the closed-form:
a
n
= a
1
· r
n−1
.
Theorem 4.2 (Sum of a Finite Geometric Sequence)
The sum of the first n terms of a geometric sequence is given by
S
n
= a
1
+ a
2
+ ··· + a
n
= a
1
·
r
n
− 1
r − 1
,
where a
1
is the first term in the sequence, and r is the common ratio.
Proof. Let
S
n
= a
1
+ a
2
+ ··· + a
n
.
Thus, the equation can be re-written as
S
n
= a
1
+ a
1
r + ··· + a
1
r
n−1
.
Multiplying both sides by r,
S
n
r = a
1
r + a
1
r
2
+ ··· + a
1
r
n
.
Subtracting the original equation from this equation,
S
n
r − S
n
= a
1
r
n
− a
1
.
Factoring S
n
out of the expression on the LHS and a
1
out of the expression on the RHS,
S
n
(r − 1) = a
1
(r
n
− 1).
Isolating S
n
,
S
n
= a
1
·
r
n
− 1
r − 1
.
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
Definition 4.3
(Infinite Geometric Sequence)
—
An
infinite
geometric sequence is a geometric sequence
with an infinite number of terms.
If the absolute value of the common ratio is less than 1, the terms will approach 0 and the sum of the terms will
approach a fixed limit. We say that the sum of the terms of this sequence is a convergent sum.
Theorem 4.4 (Sum of an Infinite Geometric Sequence)
The general formula for the sum of such a sequence is
S =
a
1
1 − r
.
Proof. Let
S = a
1
+ a
2
+ ··· .
Re-writing the equation,
S = a
1
+ a
1
r + ··· .
Multiplying both sides by r,
Sr = a
1
r + a
1
r
2
+ ··· .
Subtracting this equation from the original equation,
S − Sr = a
1
.
Factoring S out of the expression on the LHS,
S(1 − r) = a
1
.
Isolating S,
S =
a
1
(1 − r)
.
For instance, the series 1 +
1
2
+
1
4
+
1
8
+ . . ., sums to 2.
Exercise 4.5. Find the value of
1
3
+
1
9
+
1
27
+ . . .
One common instance of summing infinite geometric sequences is the decimal expansion of most rational numbers.
For instance, 0
.
33333
. . .
=
3
10
+
3
100
+
3
1000
+
3
10000
+
. . .
has first term
a
0
=
3
10
and common ratio
1
10
, so the
infinite sum has value S =
3
10
1−
1
10
=
1
3
, just as we would have expected.
Example 4.6 (AIME II 2002/11)
Two distinct, real, infinite geometric series each have a sum of 1 and have the same second term. The third
term of one of the series is 1
/
8, and the second term of both series can be written in the form
√
m−n
p
, where
m, n, and p are positive integers and m is not divisible by the square of any prime. Find 100m + 10n + p.
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
Solution.
Suppose the second term of the two distinct, real, infinite geometric series is
x.
Then, the first term is
8x
2
common ratio is
1
8x
for one of the series. Therefore, by the sum of an infinite geometric sequence,
1 =
8x
2
1 −
1
8x
=⇒ 64x
3
− 8x + 1 = 0
=⇒ (4x − 1)(16x
2
+ 4x − 1) = 0
=⇒ x =
1
4
,
−1 +
√
5
8
,
−1 −
√
5
8
.
Because the second term of both series is written in the form
√
m−n
p
,
the only applicable solution is
x
=
√
5−1
8
.
Thus, our answer is 518 .
Example 4.7 (AIME II 2011/5)
The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find
the sum of the first 6033 terms.
Solution.
The sum of the first 2011 terms is 200, second 2011 is 180, and so the ratio of the second 2011 terms
to the first 2011 terms is
9
10
. Following the same pattern, the sum of the third 2011 terms is
9
10
∗
180 = 162.
Thus, the sum of the first 6033 terms is 200 + 180 + 162 = 542 .
Example 4.8 (AIME II 2012/2)
Two geometric sequences
a
1
, a
2
, a
3
, . . .
and
b
1
, b
2
, b
3
, . . .
have the same common ratio, with
a
1
= 27,
b
1
= 99,
and a
15
= b
11
. Find a
9
.
Solution.
Let
r
be the common ratio of the two geometric sequences. Then, we see that
a
1
· r
14
=
b
1
· r
10
=
⇒
r
4
=
99
27
=
11
3
. Thus, a
9
= a
1
· r
8
= a
1
· (r
4
)
2
= 27 ·
11
3
2
= 27 ·
121
9
= 363 .
Example 4.9 (AIME I 2016/1)
For −1 < r < 1, let S(r) denote the sum of the geometric series
12 + 12r + 12r
2
+ 12r
3
+ ··· .
Let a between −1 and 1 satisfy S(a)S(−a) = 2016. Find S(a) + S(−a).
Solution. By infinite geometric series, S(a) =
12
1−a
and S(−a) =
12
1+a
. So,
S(a)S(−a) =
12
1 − a
12
1 + a
=
144
1 − a
2
= 2016.
Notice
S(a) + S(−a) =
12
1 − a
+
12
1 + a
=
24
1 − a
2
,
and so the answer is
2016
6
= 336 .
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Exercise 4.10
(AIME II 2002/3)
.
It is given that
log
6
a
+
log
6
b
+
log
6
c
= 6
,
where
a, b,
and
c
are positive
integers that form an increasing geometric sequence and b − a is the square of an integer. Find a + b + c.
Exercise 4.11
(AIME I 2009/1)
.
Call a 3-digit number geometric if it has 3 distinct digits which, when
read from left to right, form a geometric sequence. Find the difference between the largest and smallest
geometric numbers.
§5 Arithmetico-Geometric Sequence
Definition 5.1
(Arithmetico-Geometric Sequence and Series)
—
An
arithmetico-geometric series
is the
sum of consecutive terms in an arithmetico-geometric sequence defined as:
t
n
= a
n
g
n
,
where a
n
and g
n
are the nth terms of arithmetic and geometric sequences, respectively.
Example 5.2
Compute the sum
∞
X
k=1
k
4
k
.
Solution. Let the sum be S. Then
S =
1
4
+
2
4
2
+
3
4
3
+ . . . ,
and if we divide by 4,
S
4
=
1
4
2
+
2
4
3
+ . . . .
Note that if we subtract these in a special way (by subtracting the ones with common denominators), we get
3S
4
=
1
4
+
1
4
2
+
1
4
3
+ . . . =
1
4
1 −
1
4
=
1
3
,
so S =
4
9
.
Theorem 5.3 (Sum of a Finite Arithmetico-Geometric Sequence)
The sum of the first n terms of an arithmetico-geometric sequence is
a
n
g
n+1
r − 1
−
x
1
r − 1
−
d(g
n+1
− g
2
)
(r − 1)
2
,
where d is the common difference of a
n
and r is the common ratio of g
n
. We can also write it as
a
n
g
n+1
− x
1
− drS
g
r − 1
,
where S
g
is the sum of the first n terms of g
n
.
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§6 Telescoping Series
Definition 6.1
(Telescoping Series)
—
A
telescoping
series is a series whose partial sums eventually only
have a fixed number of terms after cancellation.
The cancellation technique, with part of each term canceling with part of the next term, is known as the method
of differences.
Example 6.2
Find the value of
1
2
×
2
3
×
3
4
×
4
5
×
5
6
×
6
7
.
Solution. Performing cancellations, we arrive at
1
7
.
The method of splitting the fraction is known as partial fraction decomposition.
Theorem 6.3 (Partial Fraction Decomposition)
The method is detailed below:
• Factor the denominator
• Solve the decomposition according to the right form
A few notes on decomposing:
• Linear Factors:
5x + 3
(x + 1)(x + 4)
=
A
x + 1
+
B
x + 4
• Repeated Linear Factors:
2x − 4
(x − 2)(x + 3)
2
=
A
x − 2
+
B
x + 3
+
C
(x + 3)
2
• Irreducible Quadratic Factors:
2x
2
− 3x − 1
(x − 1)(x
2
+ 9)
=
A
x − 1
+
Bx + C
x
2
+ 9
Exercise 6.4. Decompose
x
2
+ x − 5
(x − 2)(x − 1)
2
.
Partial Fraction Description is often able to be utilized in Telescoping.
Example 6.5
Find the value of
1
1 · 2
+
1
2 · 3
+
1
3 · 4
+ . . .
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Solution. If we split
1
n(n+1)
, we get
1
n
−
1
n + 1
.
Thus,
1
1
−
1
2
+
1
2
−
1
3
+
1
3
−
1
4
+ . . . = 1 .
Example 6.6
Find the value of
1 −
1
2
2
1 −
1
3
2
1 −
1
4
2
. . .
Solution. Note that
1 −
1
n
2
=
n
2
− 1
n
2
=
(n − 1)(n + 1)
n
2
.
Thus,
1 · 3
2
2
·
2 · 4
3
2
·
3 · 5
4
2
·
4 · 6
5
2
· . . . =
1
4
· 2 =
1
2
.
Example 6.7 (Stanford 2011)
Evaluate the sum
X
n≥1
7n + 32
n(n + 2)
·
3
4
n
.
Solution. Note that
7n + 32
n(n + 2)
=
16
n
−
9
n + 2
.
The sum telescopes according to the fact that 9 = (3/4)
2
· 16:
X
n≥1
16
n
3
4
n
−
16
n + 2
3
4
n+2
.
The trailing terms in the sum tend to zero as n → ∞ so the answer is
16
1
·
3
4
+
16
2
·
9
16
= 12 +
9
2
=
33
2
.
Exercise 6.8 (USAMTS 3/4/11). Determine the value of
S =
r
1 +
1
1
2
+
1
2
2
+
r
1 +
1
2
2
+
1
3
2
+ ··· +
r
1 +
1
1999
2
+
1
2000
2
Exercise 6.9
(AIME I 2002/4)
.
Consider the sequence defined by
a
k
=
1
k
2
+ k
for
k ≥
1. Given that
a
m
+ a
m+1
+ ··· + a
n−1
=
1
29
, for positive integers m and n with m < n, find m + n.
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§7 Recursive Sequences
Definition 7.1
(Recursive Sequence)
—
A
recursive sequence
is a sequence defined upon previous terms.
Example 7.2
The most famous example is the Fibonacci Sequence:
F
n
= F
n−1
+ F
n−2
,
where F
0
= F
1
= 1.
Definition 7.3
(Base Case)
—
A
base case
is one of the first few terms of a sequence, to give it an initial
value.
For example, in the sequence a
n
= a
n−1
+ a
n−2
, a
0
= a
1
= 1 would be the base cases.
Definition 7.4
(Closed Form)
—
The
closed form
, also known as the
explicit form
, of a recursive
sequence is when we do not use previous terms to define the next term.
Example 7.5
Find the closed form of
a
n
= a
n−1
+ 1,
where a
0
= 1.
Solution. The first few terms of the sequence are
a
0
= 1, a
1
= 2, a
2
= 3.
This pattern is very obviously a
n
= n + 1.
Induction
will be required to prove this solution holds forever, which again, will not be taught in this
handout.
Example 7.6 (AMC 10A 2000/6)
The Fibonacci sequence 1
,
1
,
2
,
3
,
5
,
8
,
13
,
21
, . . .
starts with two 1s, and each term afterward is the sum of its
two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the
Fibonacci sequence?
(A) 0 (B) 4 (C) 6 (D) 7 (E) 9
Solution.
Note that any digits other than the units digit will not affect the answer. So to make computation
quicker, we can just look at the Fibonacci sequence in mod 10:
1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, . . .
The last digit to appear in the units position of a number in the Fibonacci sequence is 6 =⇒ C .
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Example 7.7 (AMC 12 A 2007/25)
Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many
subsets of {1, 2, 3, . . . , 12}, including the empty set, are spacy?
Solution. Let S
n
denote the number of spacy subsets of {1, 2, 3, . . . , n}. Then, S
0
= 1, S
1
= 2, S
2
= 3.
The spacy subsets of
S
n+1
either has or does not have
n
+ 1
.
There are
S
n
spacy subsets that do not contain
n
+ 1
,
and
S
n−2
spacy subsets that do contain
n
+ 1 because if you take out
n
+ 1 from any spacy subset, the
max element is n − 2, and are possible constructions from S
n−2
.
So, we can form the recursion
S
n+1
= S
n
+ S
n−2
.
Bashing to
S
12
,
we find that the number of subsets of
{
1
,
2
,
3
, . . . ,
12
},
including the empty set, that are spacy is
129 .
Exercise 7.8.
Derive a closed form for the Fibonacci numbers defined by
F
1
=
F
2
= 1 and
F
n
=
F
n−1
+
F
n−2
.
Exercise 7.9
(AIME I 2006/11)
.
A collection of 8 cubes consists of one cube with edge-length
k
for each
integer k, 1 ≤ k ≤ 8. A tower is to be built using all 8 cubes according to the rules:
• Any cube may be the bottom cube in the tower.
• The cube immediately on top of a cube with edge-length k must have edge-length at most k + 2.
Let
T
be the number of different towers that can be constructed. What is the remainder when
T
is divided
by 1000?
For more material on recursion, view Euclid’s Orchard’s Recursion in the AMC and AIME Handout and/or
Primeri’s Recursion Handout.
§8 Problems
Problem 8.1.
Consider the arithmetic sequence with first term
−
500 and common difference 7. Find the 200th
term in the sequence.
Problem 8.2.
Consider the arithmetic sequence with first term
−
500 and common difference 7. Find
k
if
a
k
= 25.
Problem 8.3. Find the value of
1 + 2 + 3 + . . . + 100.
Problem 8.4. Find the next term of 2, 8, 32, 128, . . .
Problem 8.5. What is a
7
, if
a
n
= a
n−1
+ a
n−2
,
and a
0
= a
1
= 2?
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Problem 8.6 (AMC 10 B 2002/19). Suppose that {a
n
} is an arithmetic sequence with
a
1
+ a
2
+ ··· + a
100
= 100 and a
101
+ a
102
+ ··· + a
200
= 200.
What is the value of a
2
− a
1
?
(A) 0.0001 (B) 0.001 (C) 0.01 (D) 0.1 (E) 1
Problem 8.7
(AMC 10 B 2004/10)
.
A grocer makes a display of cans in which the top row has one can and
each lower row has two more cans than the row above it. If the display contains 100 cans, how many rows does
it contain?
(A) 5 (B) 8 (C) 9 (D) 10 (E) 11
Problem 8.8
(AMC 10 A 2006/19)
.
How many non-similar triangles have angles whose degree measures are
distinct positive integers in arithmetic progression?
(A) 0 (B) 1 (C) 59 (D) 89 (E) 178
Problem 8.9
(AMC 10 B 2016/16)
.
The sum of an infinite geometric series is a positive number
S
, and the
second term in the series is 1. What is the smallest possible value of S?
(A)
1+
√
5
2
(B) 2 (C)
√
5 (D) 3 (E) 4
Problem 8.10
(AMC 10 A 2020/21)
.
There exists a unique strictly increasing sequence of nonnegative integers
a
1
< a
2
< . . . < a
k
such that
2
289
+ 1
2
17
+ 1
= 2
a
1
+ 2
a
2
+ . . . + 2
a
k
.
What is k?
(A) 117 (B) 136 (C) 137 (D) 273 (E) 306
Problem 8.11
(AIME 1984/1)
.
Find the value of
a
2
+
a
4
+
a
6
+
a
8
+
. . .
+
a
98
if
a
1
,
a
2
,
a
3
. . .
is an arithmetic
progression with common difference 1, and a
1
+ a
2
+ a
3
+ . . . + a
98
= 137.
Problem 8.12
(AIME 1989/7)
.
If the integer
k
is added to each of the numbers 36, 300, and 596, one obtains
the squares of three consecutive terms of an arithmetic series. Find k.
Problem 8.13
(AIME II 2004/9)
.
A sequence of positive integers with
a
1
= 1 and
a
9
+
a
10
= 646 is formed so
that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic
progression, and, in general, for all
n ≥
1
,
the terms
a
2n−1
, a
2n
, a
2n+1
are in geometric progression, and the
terms
a
2n
, a
2n+1
,
and
a
2n+2
are in arithmetic progression. Let
a
n
be the greatest term in this sequence that is
less than 1000. Find n + a
n
.
Problem 8.14
(AIME II 2005/3)
.
An infinite geometric series has sum 2005. A new series, obtained by squaring
each term of the original series, has 10 times the sum of the original series. The common ratio of the original
series is
m
n
where m and n are relatively prime integers. Find m + n.
Problem 8.15
(AIME II 2016/9)
.
The sequences of positive integers 1
, a
2
, a
3
, ...
and 1
, b
2
, b
3
, ...
are an increasing
arithmetic sequence and an increasing geometric sequence, respectively. Let
c
n
=
a
n
+
b
n
. There is an integer
k
such that c
k−1
= 100 and c
k+1
= 1000. Find c
k
.
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
Problem 8.16 (Math League HS 2000-2001). Evaluate
1
2
2
+
1
3
2
+
1
4
2
+ ···
+
1
2
3
+
1
3
3
+
1
4
3
+ ···
+
1
2
4
+
1
3
4
+
1
4
4
+ ···
+ ··· .
More formally, what is the sum of all fractions of the form
1
(m+1)
(n+1)
,
where
m
and
n
range over positive
integers?
Problem 8.17 (Purple Comet HS 2004). Define a
k
= (k
2
+ 1)k! and b
k
= a
1
+ a
2
+ a
3
+ ··· + a
k
. Let
a
100
b
100
=
m
n
where m and n relatively prime natural numbers. Find n − m.
Problem 8.18
(OMO Fall 2013/24)
.
The real numbers
a
0
, a
1
, . . . , a
2013
and
b
0
, b
1
, . . . , b
2013
satisfy
a
n
=
1
63
√
2n + 2
+
a
n−1
and
b
n
=
1
96
√
2n + 2 − b
n−1
for every integer
n
= 1
,
2
, . . . ,
2013. If
a
0
=
b
2013
and
b
0
=
a
2013
,
compute
2013
X
k=1
(a
k
b
k−1
− a
k−1
b
k
) .
Problem 8.19
(Putnam 2013/B1)
.
For positive integers
n,
let the numbers
c
(
n
) be determined by the rules
c(1) = 1, c(2n) = c(n), and c(2n + 1) = (−1)
n
c(n). Find the value of
2013
X
n=1
c(n)c(n + 2).
Problem 8.20 (AoPS). If a, b, and c form an arithmetic progression, and
a = x
2
+ xy + y
2
,
b = x
2
+ xz + z
2
,
c = y
2
+ yz + z
2
,
where x + y + z 6= 0, prove that x, y, and z also form an arithmetic progression.
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freeman66 and nikenissan (February 1, 2021) Sequences and Series in the AMC and AIME
§A Appendix A: List of Theorems and Definitions
List of Theorems
2.6 Theorem - Distributive Property of Summations 4
2.8 Theorem - Commutative Property of Summations 5
3.2 Theorem - Terms of an Arithmetic Sequence 6
3.3 Theorem - Sum of an Arithmetic Sequence 6
4.2 Theorem - Sum of a Finite Geometric Sequence 8
4.4 Theorem - Sum of an Infinite Geometric Sequence 9
5.3 Theorem - Sum of a Finite Arithmetico-Geometric Sequence 11
6.3 Theorem - Partial Fraction Decomposition 12
List of Definitions
2.1 Definition - Sequence 4
2.2 Definition - Partial Sum 4
2.3 Definition - Series 4
2.4 Definition - Summation Notation 4
3.1 Definition - Arithmetic Sequence 5
4.1 Definition - Geometric Sequence 8
4.3 Definition - Infinite Geometric Sequence 9
5.1 Definition - Arithmetico-Geometric Sequence and Series 11
6.1 Definition - Telescoping Series 12
7.1 Definition - Recursive Sequence 14
7.3 Definition - Base Case 14
7.4 Definition - Closed Form 14
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