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Answer:
- a(n) = 3n -1
- a(n) = (n^2 +11)(n/3) -3
- a(n) = 3/(n+3)
- a(n) = (n+5)/11
- a(n) = n^3/(2n-1)
Step-by-step explanation:
We usually start off our study of sequences by looking at arithmetic sequences. These have a common difference. Their generic term is of the form ...
an = a1 +d(n -1)
This can be simplified, but is often left in this form so the first term (a1) and the common difference (d) are readily identifiable.
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The next sort of sequence we usually study is the geometric sequence. This has terms with a common ratio. The generic term of a geometric sequence is of the form ...
an = a1·r^(n-1)
where r is the common ratio. Again, the expression is often left in this form so the parameters are readily identifiable.
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Sequences of other kinds can be built from these, or from sequences defined by polynomial functions. One of the ways to identify the kind of sequence you have is to look at the differences of terms, and the differences of those. Differences of the given terms are called "first differences" For an arithmetic sequence, they are constant. For polynomial sequences of degrees higher than 1, the differences are constant at the level corresponding to the degree of the polynomial.
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1. First differences are constant with a value of 5-2 = 8-5 = 11-8 = 3. This is an arithmetic sequence with first term 2 and common difference 3. It can be described by ...
an = 2 +3(n -1)
In simplified form, this is ...
a(n) = 3n -1
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2. First differences are not constant, but are 6, 10, 16. Second differences are not constant, but are 10-6 = 4, 16 -10 = 6. This suggests that third differences might be constant with a value of 6 -4 = 2. In any event, a sequence of 4 terms can always be matched by a 3rd-degree polynomial.* Here, that polynomial would be ...
an = 1 +(n -1)(6 +(n -2)/2(4 +(n -3)/3(2)))
a(n) = (n^2 +11)(n/3) -3
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3. We recognize this as a harmonic sequence with 3 in the numerator and denominators being an arithmetic sequence increasing by 1 starting with 4.
a(n) = 3/(n +3)
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4. Here, the denominator is constant and the numerator increases by 1 starting with 6.
a(n) = (n +5)/11
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5. We recognize the numerators are the sequence of cubes, and the denominators are an arithmetic sequence starting with 1 and increasing by 2.
a(n) = n^3/(2n -1)
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* In general, you can find the coefficients of a polynomial sequence by solving the equations for those coefficients. Here, that would be 4 linear equations in 4 unknown values. It is tedious, but not difficult.
Once you know the set of differences down to the level where they are constant, you can write the polynomial using a generic form. If dn is the first of the differences at level n, then the terms become ...
an = a1 +(n-1)d1/1! +(n-1)(n-2)d2/2! +(n -1)(n -2)(n -3)d3/3! +...
In the above, we have factored this sequence into "Horner Form" using factors of the form (n -k)/k. It is easier to write, but perhaps more tedious to simplify.
