# Finding the general term of a sequence

I would like to find an expression for the sequence $\;\{a_n\},\;n=0,\,1,\,2,\,3,\,\dots,\;$ $$-\frac{1}{6},\,\frac{2}{7},\,\frac{5}{8},\,\frac{8}{9},\,\frac{11}{10},\,\ldots$$

So by trial and error I got $$a_n\,=\,\frac{3n-1}{n+6},\; n=0,\,1,\,2,\,\ldots$$

How does one approach these types of problems in general and in particular, the present problem, other than by trial and error?

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$$-\frac{1}{6},\,\frac{2}{7},\,\frac{5}{8},\,\frac{8}{9},\,\frac{11}{10},\,\ldots \quad a_n = \;\;?$$

Look for patterns:

• Numerator: each term increases by $3$, starting at $-1$ when $n = 0\;\; \rightarrow \;\;3\cdot n - 1\;$?

• Denominator: each term increases by $1$, starting at $6$ when $n = 0 \;\; \rightarrow \;\;1\cdot n+6\;$?

Note that both the numerators and the denominators form arithmetic progressions.

Test it:

$$\{a_n\},\; n\in \{0, 1, 2, 3, ...\};\;\;a_n\;=\; \dfrac{3n - 1}{n+6} \quad = \quad -\frac{1}{6},\,\frac{2}{7},\,\frac{5}{8},\,\frac{8}{9},\,\frac{11}{10},\,\frac{14}{11},\, \frac{17}{12},\,\ldots$$

If testing it fails, try to "tweak it" the expression of the general term (reevaluate to ensure correct term for $a_0$, ensure that the numerator and denominators do, in fact, represent arithmetic progressions...etc.)

Not applicable to your particular sequence, but a useful tool to have in your toolbox of strategies: Another progression you'll see arise frequently in sequences is the geometric progression, another "good to know" progression to learn so you can spot it when it occurs, and make use of how to form the general expression for a term in such a sequence.

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So if the sequence starts from zero and it is an AP, would the general term be $a_n=a_0 +nd$? – Gorg Jan 24 '13 at 16:58
Yes! I think you've got it: $a_0$ being the term at n = 0. – amWhy Jan 24 '13 at 16:59
Thanks very much. – Gorg Jan 24 '13 at 17:04
Glad to help, Gorg! – amWhy Jan 24 '13 at 17:05