Arithmetic Sequences

**KittyOfDoom** · April 23rd 2008, 04:03 PM

I have been having trouble with the following problem:
Write the next term in the sequence. Then write a rule for the nth term.

7,-8,9,-10... I found out that the next number in the sequence is 11, but I have yet to figure out the rule for the nth term. I have been trying to solve this problem for a while and cannot. Can you help?

**Mathstud28** · April 23rd 2008, 04:22 PM

Originally Posted by KittyOfDoom

I have been having trouble with the following problem:
Write the next term in the sequence. Then write a rule for the nth term.

7,-8,9,-10... I found out that the next number in the sequence is 11, but I have yet to figure out the rule for the nth term. I have been trying to solve this problem for a while and cannot. Can you help?

$a_n=(n+6)(-1)^{n+1}$

**Soroban** · April 23rd 2008, 04:22 PM

Hello, KittyOfDoom!

Write the next term in the sequence. Then write a rule for the nth term.
. . . $7,\:-8,\:9,\:-10,\:\hdots$

Ignoring the signs, we have: . $7,\:8,\:9,\;10,\:\hdots$

. . The rule (so far) is: . $f(n) \:=\:n+6$

Now, how do we make the signs alternate?
. . If you've never seen it done, this can be baffling.

Consider $(-1)^n$ . . . negative-one raised to consecutive powers.

We get:
. . . . . . . $\begin{array}{ccc}(-1)^1 &=& -1 \\ (-1)^2 &=& +1 \\ (-1)^3 &=& -1 \\ (-1)^4 &=& +1 \\ (-1)^5 &=&-1 \\ \vdots & & \vdots \end{array}\quad\hdots$ . See? .Alternating signs!

If we want to start with "minus", we use: . $(-1)^n$

If we want to start with "plus", we use: . $(-1)^{n+1}$

For this problem, the rule is: . $\boxed{f(n) \;=\;(-1)^{n+1}(n+6)}$

. . Plug in $n = 1,2,3,\hdots$ and check it out.

**KittyOfDoom** · April 23rd 2008, 04:25 PM

Thank you for the help ^.^

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