Sequence and series
Sequence and Series
A "sequence" or progression is an planned list of numbers; the numbers in this ordered list are referred to as "elements" or "terms". A "series" is the value you obtain when you add up all the terms of a sequence; this value is known as the "sum". For example, "1, 2, 3, 4" is a sequence, with terms "1", "2", "3", and "4"; the equivalent series is the sum "1 + 2 + 3 + 4", and the value of the series is 10.
A sequence may be named or known as "A" or "An". The terms of a sequence are normally named something like "ai" or "an", with the subscripted letter "i" or "n" being the "index" or counter. Thus the second term of a sequence may be referred "a2" (pronounced "ay-sub-two"), and "a12" would assign the twelfth term.
Note: Sometimes sequences start with an index of n = 0, so the first term is actually a0. Then the second term would be a1. The first listed term in such a case would be known as the "zero-eth" term. This method of numbering the terms is used, for instance, in Javascript arrays. Don't assume that every sequence and series will begin with an index of n = 1.
A sequence A with terms an may as well be known as "{an}", but contrary to what you may have learned in other contexts, this "set" is in fact an ordered list, not an unordered collection of elements. Unfortunately, notation doesn't yet appear to have been completely standardized for this topic. Just try constantly make sure, whatever resource you're making use of that you are clear on the definitions of that resource's terms and symbols.
To designate a series, we make use of either the Latin capital letter "S" or else the Greek letter corresponding to the capital "S", which is is known as "sigma" (SIGG-muh):
To illustrate the summation of, for example, the first through tenth terms of a sequence {an}, we would write the following:
The "n = 1" is the "lower index", informing us that "n" is the counter and that the counter begins at "1"; the "10" is the "upper index", informing us that a10 will be the last term added in this series; "an" stands for the terms that we'll be adding. The whole thing is pronounced as "the sum, from n equals one to ten, of a-sub-n". The summation symbol above means the following:
a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 + a9 + a10
The written-out form above is referred to as the "expanded" form of the series, contrary to the more compact "sigma" notation.
Any letter can be made use of for the index, but i, j, k, and n are likely used more than any other letters.
Sequences and series are most cooperative when there is a formula for their terms. For example, if the formula for an is "2n + 3", then you can obtain the value of any term by plugging the value of n into the formula. For example, a8 = 2(8) + 3 = 16 + 3 = 19.
This means that "an = 2n + 3" can be read as "the n-th term is given by two-enn plus three". The word "n-th" is pronounced "ENN-eth", and merely means "the generic term an, where we haven't already specified the value of n."
Of course, there doesn't need to be a formula for the n-th term of a sequence. The values of the terms can be completely random, having no relationship between n and the value of an. But sequences with arbitrary terms are hard to work with and are generally less helpful, therefore you're not likely to encounter a lot of them in your classes.
• Let An = {1, 3, 5, 7, 9}.
What is the value of a3?
Find the value of
The index of a3 is n = 3, therefore, what you are being asked to obtain for the third term, which is "5". The "value" that is required is the total, the sum, of all the terms an from a1 to a5; in other words:
a1 + a2 + a3 + a4 + a5 = 1 + 3 + 5 + 7 + 9 = 25
value of a3: 5 value of sum: 25
Expand the following series and find the sum:
To obtain each term, you ought to input the value of n into the formula. In this instance, you would be beginning with n = 0 and ending with n = 4.
2(0) + 2(1) + 2(2) + 2(3) + 2(4) = 0 + 2 + 4 + 6 + 8 = 20
• List the first four terms of the sequence {an} = {n2}, starting with n = 1.
You will merely plug n into the formula, and simplify:
{a1, a2, a3, a4} = {12, 22, 32, 42} = {1, 4, 9, 16}
• List the first four terms of the sequence below, starting with n = 0:
A lot of sequences and series are made up of factorials, and this is one of them. You would be required to calculate it the normal way:
a0=(-1)0⁄(0 + 1)!=1⁄1!=1⁄1=1
a1=(-1)1⁄(1 + 1)!=-1⁄2!=1⁄2
a2=(-1)2⁄(2 + 1)!=1⁄3!=1⁄6
a3=(-1)3⁄(3 + 1)!=-1⁄4!=1⁄24
Therefore the terms are:
a0,a1,a2,a3=1,1⁄2,1⁄6,1⁄24
Observe the way, in that last instance above, raising the –1 to the power n made the signs swap. This swapping pattern of signs is seen a lot of times especially in calculus, thus try to keep this "raising –1 to the power n" secret in mind.
Find the sum of the first six terms of An, where an = 2an–1 + an–2, a1 = 1, and a2 = 1.
This formula appears much worse than it actually is; you merely have to give yourself some time, and dissect the formula cautiously.
They provided me the values of the first two terms, and then they gave me a formula that says that every term (after the first two terms) is a sum formed from the previous two terms. Putting it into the formula, you will obtain:
a3 = 2a3–1 + a3–2 = 2a2 + a1 = 2(1) + (1) = 2 + 1 = 3
a4 = 2a4–1 + a4–2 = 2a3 + a2 = 2(3) + (1) = 6 + 1 = 7
a5 = 2a5–1 + a5–2 = 2a4 + a3 = 2(7) + (3) = 14 + 3 = 17
a6 = 2a6–1 + a6–2 = 2a5 + a4 = 2(17) + (7) = 34 + 7 = 41
Now that you've obtained the values of the third through the sixth terms, you can find the value of the series; the sum is:
1 + 1 + 3 + 7 + 17 + 41 = 70
• Write the following series with the use of summation notation, starting with n = 1:
2 – 4 + 6 – 8 + 10
The first thing we have to do is find out a relationship between n and the terms in the summation. This series is more or less simple, though: each term an is twice n, therefore, there is clearly a "2n" in the formula. You as well have the swapping sign. If you make use of (–1)n, you'll obtain –2, 4, –6, 8, –10, which is backwards (on the signs) from what you want. But you can swap the signs by throwing in one more factor of –1:
(–1)(–1)n = (–1)1(–1)n = (–1)n+1
Therefore, the formula for the n-th term is an = (–1)n+1(2n). For the fact that n begins at 1 and there are five terms, then the summation is:
• Write the following with the use of summation notation:
5⁄6 + 3 + 5⁄7 + 3 + 5⁄8 + 3+ · · ·+5⁄31 + 3
The only thing that alters from one term to the other is one of the numbers in the denominator. (If you "simplify" these fractions, you'll lose this information. Any time the terms of your sequence or series look strangely lumpy, you tend not to simplify those terms: that odd lumpiness nearly certainly is made up of a hint of the pattern you are supposed to find.) The alternating numbers, as a list, are 6, 7, and 8. This is very much similar to counting, but beginning with 6 rather than 1. Without any information to the contrary, you'll presume that this is the pattern.
But you are required to connect to these "counting" values to the counter, the index, n. For n = 1, the number is 6, or n + 5. For n = 2, the number is 7, which is also n + 5. Calculating the pattern for n = 3, 3 + 5 = 8, which is the third number. Then the terms appears to be in the pattern below:
5⁄(n + 5)3
But how do you know the number of terms that are in the summation? The ellipsis (the "..." or "dot, dot, dot" in the middle) is used to indicate that terms were omitted. Nevertheless, now that we have the general pattern for the series terms, we can solve for the counter (the value of n) in the last term:
31 = n + 5
31 – 5 = n + 5 – 5
n = 26
This signifies to us that there are 26 terms in this summation, therefore the series, in summation notation, is:
If the fractions (above) had been simplified and reduced, it would have been a lot harder to figure out a pattern. Unless the sequence is very simple or is presented in a very straightforward manner, it is possible that you won't be able to find a pattern, or might find a "wrong" pattern. Don't let this bother you terribly much: the "right" pattern is just the one that the author had in mind when he wrote the problem. Your pattern would be "wrong" only in that it is unexpected. But if you can present your work sensibly and mathematically, you ought to able get to the point that you would earn at least partial credit for your answer.
Immediately you've learned the basic notation and terminology, the two simplest sequences to work with are arithmetic and geometric sequences. An arithmetic sequence moves from one term to the next by continually adding (or subtracting) an equivalent value. For example, 2, 5, 8, 11, 14,... and 7, 3, –1, –5,... are arithmetic, due to the fact that you add 3 and subtract 4, correspondingly, at every step. A geometric sequence moves from one term to the next by constantly multiplying (or dividing) by the equivalent value. Thus, 1, 2, 4, 8, 16,... and 81, 27, 9, 3, 1, 1/3,... are geometric, due to the fact that you multiply by 2 and divide by 3, correspondingly, at every step.
The number added (or subtracted) at every stage of an arithmetic sequence is known as the "common difference" d, due to the fact that if you subtract (obtain the difference of) successive terms, you'll at all time obtain this common value. The number multiplied (or divided) at every stage of a geometric sequence is known as the "common ratio" r, due to the fact that if you divide (find the ratio of) successive terms, you'll constantly obtain this common value.
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