# Finding a general term for the sequence

Find a general term in simplest form for the sequence: $$2, 1, -4, 7, -10, 13, -16$$

This is what I tried:

$a_n = a_1 + (n - 1)\cdot d$
$a_n = 2 + (n-1)\cdot 3$
$a_n = (-1)^n (2+(n-1)\cdot 3)$

which doesn't work.

I am pretty confused regarding the first two terms of the sequence and how to deal with the lack of a negative number (all of the terms after that seem to follow a pattern of +,-,+,- so I think multiplying by $(-1)^n$ would help there), as well as the different change in value from the first to the second term in the sequence, where as all the following numbers seem to follow a more recognizable pattern to me.

Any help would be greatly appreciated.

• You're making it too complicated. Hint: look at every other term. As for the starting terms, simply define them to be $a_1 = 2, a_2 = 1$. – Théophile Jul 30 '15 at 23:37
• The clearest way to write the general term, in my opinion, is in the form: $$a_n = \begin{cases} \ldots & \text{if n is odd} \\ \ldots & \text{if n is even} \end{cases}$$ – Théophile Jul 30 '15 at 23:42
• Hmm. Alright; thank you for the help – 121c Jul 30 '15 at 23:46

Let $\{a_n\}$ be the sequence we want.

Odd terms $\{b_n\}$ : $2,-4,-10,-16,\cdots$ imply that $b_n=-6n+8$.

Even terms $\{c_n\}$ : $1,7,13,\cdots$ imply that $c_n=6n-5$.

So, we have $a_{2n-1}=b_n=-6n+8$ and $a_{2n}=c_n=6n-5$.

So, for $N=2n-1$ odd, $$a_N=-6\cdot\frac{N+1}{2}+8=-3N+5=-(3N-5)=(-1)^N(3N-5).$$ And for $N=2n$ even, $$a_N=6\cdot\frac N2-5=3N-5=(-1)^N(3N-5).$$

Thus, we have $a_n=\color{red}{(-1)^{n}(3n-5)}\ (n=1,2,\cdots)$.

• Oh okay. I see. Thank you for the help. Cheers – 121c Jul 30 '15 at 23:55