how to solve sequence problems
What does that have to do wth geometry? Can you give an example of a sequence problem? If you are looking for the sum or the limit of a sequence, different methods are used depending upon the sequence.i
Number sequences of many varieties have been the focus of study and enjoyment since the ten digits first appeared on the scene. Our first exposure to number sequences was, in fact, the counting numbers, or natural numbers, that we learned in our first year of school. WIthout question, the sequence of counting numbers is the most fundamental of number sequences and all other number sequences derive from manipulations of these counting numbers. A broad, in depth, discussion of sequences would take up too much space here and, therefore, the following is but a brief introduction to the topic. It is my hope that the material that follows will be interesting and entertaining enough to spur you to seek out further information, and believe me, there is much more to be learned. Lets see where these magical numbers take us.
> The Natural Numbers are the familiar set of whole numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,.....etc., that we see and use every day. (The 0 is sometimes included.) The set of natural numbers is often referred to as the counting numbers and often denoted by N.
> A number sequence, or progression, is an orderly set of numbers arranged in such a way that each
successive number in the sequence is defined by a fixed rule or law related to the position in the sequence or the previous number, or numbers, in the sequence. The specific numbers in the sequence are called the terms or elements of the sequence.
> A number series is the sum of the terms of a sequence.
> The sum of the numbers in a sequence, or the series of numbers, can be finite or infinite. Finite sequences and series have a well defined first and last term while infinite sequences and series continue forever. A finite set of numbers is typically portrayed as [1, 4, 7, 10, 13, 16] while an infinite set of numbers is usually portrayed as [1, 3, 5, 7, 9, ......]. Very often, the brackets are omitted.
> A divergent sequence/series is one having no finite sum.
> A convergent sequence/series is one having a finite sum or limit.
> Sequences/Series come in many varieties, arithmetic, geometric, harmonic, power, finite difference, binary, triad, to name a few, plus many known primarily by the name of their discoverer such as Fibonacci, Lucas, Pell, and Pascal. Others derive their designations from specific geometric manipulations or arrangements of the counting numbers such as figurate or polygonal, pyramidal, tetrahedral, etc. Lets explore.
Elementary Sequences
The most elementary sequence we are familiar with is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,..etc., or the counting numbers.
Removing every other number starting with the 2, or the even numbers, leaves us with another sequence
1, 3, 5, 7, 9, 11, 13, 15,.... or the sequence of odd numbers. The nth term of this sequence is given by N(n) = 2n - 1.
Take the successive sums of these odd numbers to create the sequence
1, 4, 9, 16, 25, 36, 49, 64, 81, 100....etc.
Look familiar? The sequence of perfect squares.
The sum of n terms of the odd number sequence is therefore given by S(n) = n^2.
Consider the sequence of triangular numbers, the successive sums of the counting numbers, i.e., 1, 1+2=3, 1+2+3=6, 1+2+3+4=10, etc., or
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, etc.
To these add the same sequence of numbers staring with the 2nd triangular number as in
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, etc
.....1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, etc
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, etc.
How elegant, the sequence of perfect squares again.
Starting with the counting numbers again:
Remove every third counting number starting with the 3 which leaves us with
1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, etc.
Take the successive sums of these numbers to create the sequence
1, 3, 7, 12, 19, 27, 37, 48, 61, 75, 91, 108, 127, 147, 169, 192, etc.
Remove every other number starting with the 3 which leaves us with
1, 7, 19, 37, 61, 91, 127, 169, etc.
Take the successive sums of these numbers to create the sequence
1, 8, 27, 64, 125, 216, 343, 512, etc.
Look familiar? The sequence of perfect cubes.
Another relationship between triangular numbers and their squares.
...Triangular....1.......3.......6.......10.......15.......21........28
...Square........1..... .9......36.....100......225.....441......784
...Difference.......8......27......64......125.....216......343
...Equals..........2^3....3^3.....4^3......5^3......6^3......7^3
Thus, from the nth Triangular number, Tn, (Tn)^2 - (T(n-1))^2 = n^3.
The possibilities are many and, unfortunately, too numerous to fully present here, but you can readily see how the basic counting numbers can be manipulated to create many unique sequences.
Lets explore some of the more familiar sequences that we come in contact with.
Arithmetic Sequences
A set of numbers are said to form an arithmetic sequence or arithmetic progression when they increase or decrease by a common difference, as in a, (a + d), (a + 2d), (a + 3d), --------(a + nd), "a" being the first term, "d" being the common difference, and "n" being the number of terms in the sequence.
The last, or nth, term of an arithmetic progression is L = a + (n - 1)d.
The sum of an arithmetic sequence/progression is given by S = n(a + L)/2.
The arithmetic mean of three quantities in arithmetic progression is the middle term.
The arithmetic mean between any two quantities is Am = (a + b)/2.
The sum of the set of consecutive positive integers beginning with 1 is S = n(n + 1)/2.
The sum of a set of consecutive positive integers from n1 to n2 is S = [n2(n2 - 1) - n1(n1 - 1)]/2.
> The product of four consecutive numbers of an arithmetic progression of integers plus the 4th power of the common difference is always a perfect square. Example: From 3, 6, 9, 12, we derive (3x6x9x12) + 3^4 = 1944 + 81 = 2025 = 45^2.
Geometric Sequences
Numbers are said to form a geometric sequence or geometric progression when they increase or decrease by a constant factor as in a, ar, ar^2, ar^3,........ar^n, "a" being the first term, "r" being the common factor, and "n" being the number of terms. The constant factor is also called the common ratio..
The last term of a geometric progression is given by L = ar^(n - 1).
The sum of a finite geometric progression is given by S = a[(r^n) - 1]/(r - 1).
The geometric mean of three quantities in geometric progression is the middle term.
The geometric mean between any two separated terms is Gm = sqrt(ab).
The sum of an infinite geometric progression is infinite when the common ratio, r, is greater than 1.
The sum of an infinite geometric progression is finite and given by Sn = a/(1-r) when -1 < r < +1.
Harmonic Sequences
Three quantities are said to be in harmonic progression when a/c = (a - b)/(b - c). Any number of quantities are said to be in harmonic progression when every three consecutive terms are in harmonic progression. The reciprocals of quantities in harmonic progression are in arithmetic progression. If a, b, and c are in harmonic progression, i.e., a/c = (a - b)/(b - c), a(b - c) = c(a - b). Dividing every term by abc yields 1/c - 1/b = 1/b - 1/a, thus proving the proposition. The nth term of the harmonic progression 1/a + 1/b + 1/c + 1/d +......., is given by H(n) = 1/[a + (n-1)(b-a)].
The sequence of terms derived from the reciprocals of an arithmetic progression form a harmonic progression as in 1, 1/2, 1/3, 1/4, ...........1/n. Problems in harmonic progressions are sometimes solved by inverting the terms, and treating the result as an arithmetic progression.
The harmonic mean of three quantities in harmonic progression is the middle term. The harmonic mean between any two quantities is given by Hm = 2ab/(a + b).
Surprisingly, the harmonic series diverges.
The sum of the harmonic series 1/1 + 1/2 + 1/3 + 1/4 +...........1/n is defined by Sn = ln(n) + .577215664901, the number .577215664901.......being Euler's constant.
If A, G, and H are the arithmetic, geometric, and harmonic means between a and b, from A = (a + b)/2, G = sqrt(ab), and H = 2ab/(a + b), we can derive AH = G^2, meaning that G is the geometric mean between A and H.
Finite Difference Sequences
A finite difference sequence is one where the successive differences between the terms eventually result in a constant. For example, consider the 1 through n terms N1, N2, N3, N4,........Nn:
n...................1......2......3......4......5......6......7.......8.......9......10
N..................2......6.....13.....23....36....52....71....93....118....136
1st Diff.............4......7......10....13....16....19....22....25......28
2nd Diff................3......3.......3......3......3......3......3.......3
As with most sequences, an expression can be derived enabling the definition of the nth term of a finite difference sequence. The expression is a function of the number of successive differences required to reach the constant difference. If the first differences are constant, the expression is of the first order, i.e., N = an + b. If the second differences are constant, the expression is of the second order, i.e., N = an^2 + bn + c. Similarly, constant third differences derive from N = an^3 + bn^2 + cn + d. Take the following example:
n.................1......2......3......4......5......6................n
N................3......9.....19.....35....59....93...............Nn
1st Diff............6.....10.....16....24....34
2nd Diff...............4.......6......8.....10
3rd Diff....................2......2......2
Using the data points (n1, N1), (n2,N2), (n3,N3), etc., we substitute them into N = an^3 + bn^2 + cn + d as follows:
(n1,N1) = (1,3) produces a(1^3) + b(1^2) + c(1) + d = 3 or a + b + c + d = 3
(n2,N2) = (2,9) produces a(2^3) + b(2^2) + c(2) + d = 9 or 8a + 4b + 2c + d = 9
(n3,N3) = (3,19) produces a(3^3) + b(3^2) + c(3) + d = 19 or 27a + 9b + 3c + d = 19
(n4,N4) = (4,35) produces a(4^3) + b(4^2) + c(4) + d = 35 or 64a + 16b + 4c + d = 35
Subtracting each successive pair yields
7a + 3b + c = 6
19a + 5b + c = 10
37a + 7b + c = 16
Again, subtracting each successive pair yields
12a + 2b = 4
18a + 2b = 6
Subtracting these yields 6a = 2 making a = 1/3, b = 0, c = 11/3, and d = -1 resulting in our final expression for the nth term of this sequence Nn = (n^3)/3 + (11n)/3 - 1. Checking it out for the 6th term we have (6^3)/3 + (66)/3 - 1 = 72 + 22 - 1 = 93.
Another way of looking at the finite difference sequence is as follows:
Assume x(n) represents a rational integral sequence of n terms, denoted by x1, x2, x3, x4-----x(n). By subtracting successive terms, we can obtain another sequence. Denote this first order set of differences as d1x1, d1x2, d1x3, d1x4.......etc. We can now subtract each term of the first order differences and create another sequence, the second order set of differences, denoted by d2x1, d2x2, d2x3, d2x4.......etc. Obviously, we can continue to form further sequences of the 3rd, 4th, 5th,.....orders of differences. Eventually, the differences will be a constant. Without getting into the literal algebraic derivation, a sequence of this type can be defined by
x(n) = x1 + (n - 1)d1 + [(n - 1)(n - 2)d2]/1(2) + [(n - 1)(n - 2)(n - 3)d3]/1(2)3 + ......etc.
If we denote the first term of a sequence of n terms as "a" with the sequence d1, d2, d3, ....as the first terms of the successive orders of differences, any term of the sequence may be derived from
a(n) = a + (n - 1)d1 + [(n - 1)(n - 2)d2]/2! + [(n - 1)(n - 2)(n - 3)d3]/3! + .....with ! meaning factorial.
The sum of n terms of the sequence is given by
S(n) = na + [n(n - 1)d1]/2! + [n(n - 1)(n - 2)d2]/3! + [n(n - 1)(n - 2)(n - 3)d3]/4! + ......
To repeat, this definition and summation is applicable to sequences where, eventually, in one of the difference sequences, the terms are all equal. An example will serve to illustrate.
Suppose you were asked to erect a monument in the form of a pyramidal pile of cannonballs with a square base that was 5 cannonballs high. You had to go get the correct amount of cannonballs from a supply house. How many would you buy?
What does a square based pyramidal ppile of cannonballs look like? One ball on top. The next layer contains 4 balls. The next 9 balls. The next 25 balls. What does this series look like?
......................1^1....2^2....3^2....4^2.....5^2....6^2
Series..............1.......4.......9......16......25.....36
1st differences......3.......5........7.......9.......11.......d1
2nd differences.........2........2........2......2..............d2 = a constant
Lets apply our two expressions to verify the 6th term and the sum of the first 6 terms.
For our seires, a = 1, n = 6, d1 = 3, and d2 = 2. So, we have
a(6) = 1 + (6 - 1)3 + [(6 - 1)(6 - 2)2]/2 = 1 + 15 + 20 = 36 = 6^2.
The sum of these 6 terms ending with 36 becomes
S(6) = 6(1) + [6(6 - 1)3]/2 + [6(6 - 1)(6 - 2)2]/6 = 6 + 45 + 40 = 91.
Binary Sequence
The binary sequence is the sequence of positive powers of 2. The sequence is simply 1, 2, 4, 8, 16, 32, 64, 128, 256, etc. It is worth noting that the terms of this sequence can be combined to create any of the counting numbers. This unique property is often portrayed through Bachet's famous problem regarding finding the least number of weights which would enable the weighing of any integral number of pounds from 1 to 40 lb. inclusive. Bachet presented two solutions, one of which was derived in the year 1556, namely the series of weights 1, 2, 4, 8, 16, and 32 lb. This solution assumes that the weights were placed on one side of the balance scale only, and thereby actually enabled the weighing of up to 63 lb. Bachet also took another approach as explained in the Triad sequence below.
The 3rd term of harmonic progression is 15 and the 9th term is 6. Find the 11th term
Least number of weights to weigh any integral weight from 1 to 40 is 4 according to me. The weights would be-1,3,9 and 27 and may be placed on either side.
To solve sequence problems, follow these steps:
1. Identify the pattern: Look for any consistent changes or repetitions in the given sequence. Pay attention to the differences between consecutive terms or any common ratios.
2. Determine the formula: Once you have identified the pattern, try to find a formula or equation that can represent the sequence. This formula will allow you to calculate any term in the sequence.
3. Find the missing term: If you are given a sequence with one or more missing terms, use the formula you found in step 2 to calculate the missing term(s).
4. Check for errors: After finding the missing term, double-check your work by recalculating a few terms of the sequence to ensure that your formula is correct.
Here are a few common types of sequence problems, along with strategies to solve them:
1. Arithmetic Sequences: In an arithmetic sequence, each term is obtained by adding a constant difference to the previous term. To find any term in an arithmetic sequence, use the formula: a_n = a_1 + (n-1)d, where a_n is the nth term, a_1 is the first term, n is the term number, and d is the common difference.
2. Geometric Sequences: In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio. To find any term in a geometric sequence, use the formula: a_n = a_1 * r^(n-1), where a_n is the nth term, a_1 is the first term, n is the term number, and r is the common ratio.
3. Recursive Sequences: In a recursive sequence, each term is determined by previous terms in the sequence using a given rule. To find any term in a recursive sequence, you may need to use a combination of arithmetic or geometric sequences and mathematical operations to derive the formula.
Remember, practice is key to improving your skills in solving sequence problems. So, try to solve various sequence problems to familiarize yourself with different patterns and techniques.