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Sequences and Series: Basic Examples (page 2 of 5) Sections: Terminology and notation, Basic examples, Arithmetic and geometric sequences, Arithmetic series, Finite and infinite geometric series
The index of a_{3} is n = 3, so they're asking me for the third term, which is "5". The "value" they're asking for is the total, the sum, of all the terms a_{n} from a_{1} to a_{5}; in other words: a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 1 + 3 + 5 + 7 + 9 = 25 value of a_{3}:
5
To find each term, I'll plug the value
of n into
the formula. In this case, I'll be starting with 2(0) + 2(1) + 2(2) + 2(3) + 2(4) = 0 + 2 + 4 + 6 + 8 = 20
I'll just plug n into the formula, and simplify: {a_{1}, a_{2}, a_{3}, a_{4}} = {1^{2}, 2^{2}, 3^{2}, 4^{2}} = {1, 4, 9, 16}
Many sequences and series contain factorials, and this is one of them. I'll evaluate in the usual way: So the terms are: Copyright © Elizabeth Stapel 20062011 All Rights Reserved Notice how, in that last example above, raising the –1 to the power n made the signs alternate. This alternating pattern of signs crops up a lot, especially in calculus, so try to keep this "raising –1 to the power n" trick in mind.
This formula looks much worse than it really is; I just have to give myself some time, and dissect the formula carefully. They gave me the values of the first two terms, and then they gave me a formula that says that each term (after the first two terms) is a sum formed from the previous two terms. Plugging into the formula, I get: a_{3} = 2a_{3–1}
+ a_{3–2} = 2a_{2} + a_{1}
= 2(1) + (1) = 2 + 1 = 3
Now that I've found the values of the third through the sixth terms, I can find the value of the series; the sum is: 1 + 1 + 3 + 7 + 17 + 41 = 70
2 – 4 + 6 – 8 + 10 The first thing I have to do is figure out a relationship between n and the terms in the summation. This series is pretty easy, though: each term a_{n} is twice n, so there is clearly a "2n" in the formula. I also have the alternating sign. If I use (–1)^{n}, I'll get –2, 4, –6, 8, –10, which is backwards (on the signs) from what I want. But I can switch the signs by throwing in one more factor of –1: (–1)(–1)^{n} = (–1)^{1}(–1)^{n} = (–1)^{n}^{+1} So the formula for the nth term is a_{n} = (–1)^{n}^{+1}(2n). Since n starts at 1 and there are five terms, then the summation is:
The only thing that changes from one term to the next is one of the numbers in the denominator. (If I "simplify" these fractions, I'll lose this information. Any time the terms of my sequence or series look oddly lumpy, I tend not to simplify those terms: that odd lumpiness almost certainly contains a hint of the pattern I need to find.) The changing numbers, as a list, are 6, 7, and 8. This looks like counting, but starting with 6 instead of 1. Without any information to the contrary, I'll assume that this is the pattern.
But I need to relate 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. Checking the pattern for n = 3, 3 + 5 = 8, which is the third number. Then the terms seems to be in the following pattern: But how many terms are in the summation? The ellipsis (the "..." or "dot, dot, dot" in the middle) means that terms were omitted. However, now that I have the general pattern for the series terms, I can solve for the counter (the value of n) in the last term: 31 = n + 5
This tells me that there are 26 terms in this summation, so 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 should be able to talk your way into getting at least partial credit for your answer. Once you've learned the basic notation and terminology, you should quickly move on to the two common and straightforward sequence types.... << Previous Top  1  2  3  4  5  Return to Index Next >>



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