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Find the $5^{th}$ and $8^{th}$ terms of the sequence defined by

$$a_n=\begin{cases} n(n+1),&\text{if }n\text{ is an even number, and}\\\\ \frac{n}{n^2-1},&\text{if }n\text{ is an odd number}\;. \end{cases}$$

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closed as too localized by Andres Caicedo, tomasz, Jason DeVito, userNaN, Noah Snyder Oct 13 '12 at 21:53

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Since $a_n$ depends only on $n$ you just need to substitute $n$ into the relevant formula, depending on whether $n$ is even or odd. –  Mark Bennet Oct 12 '12 at 17:43
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The only possible question is whether the $5$-th term is $a_5$ or, if your indexing starts at $0$, $a_4$. In any case, just substitute the appropriate values of $n$ into the formula for $a_n$. –  Brian M. Scott Oct 12 '12 at 17:46
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Except, Bryan, if the indexing starts at $0$, then we must skip the index $n=1$, so it ends up being the same (if we don't skip any others). –  Cameron Buie Oct 12 '12 at 17:56

1 Answer 1

up vote 1 down vote accepted

There are a few issues that need to be addressed, here. First, the more general issue of which index we are taking to be the "$n$th" term of the sequence. If we mean "the term corresponding to the $n$th least index in the domain that we chose for the sequence" (remember, a sequence is simply a function on a countable set), then it depends on the domain we chose. Moreover, if this is the case, then note that we can't have our index set be the natural numbers--after all, what is $a_1$?

Now, if by "$n$th" term, we mean the term with index $n$, then this is simply a matter of evaluation (so I think that's probably what is intended, since no specific index set was given, so far as you've mentioned). Specifically, we'll be "plugging in" $n=5$ and $n=8$ with the appropriate formula in each case.

For example, if we want the 4th term in the sequence--that is, $a_4$--then since $4$ is even we have $$a_4=4(4+1)=4\cdot 5=20.$$ If we want the $7$th term--$a_7$--then since $7$ is odd, we have $$a_7=\frac{7}{7^2-1}=\frac7{48}.$$

Does that give you the idea?

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Thank You for the help. Got the point. –  Alpha Oct 13 '12 at 3:50

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