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.
 A: 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 $a_n=\color{red}{(-1)^{n}(3n-5)}\ (n=1,2,\cdots)$.
A: There is a pattern in this sequence:$$\frac{a_1 + a_3}{2}=-a_2,\space \space \space \space \frac{a_2+a_4}{2}=-a_3,\space \space \space \space \frac{a_3+a_5}{2}=-a_4$$ Using mathematical induction, it can be deduced that: $$a_{n}+2a_{n+1}+a_{n+2}=0$$ This is indeed a good deduction but it won't help us find the general term of the sequence alone. Hence, we need to go further:

*

*$a_{n+1}-a_{n}=b_{n}, \space \space \space \space b_{n+1}-b_{n}=c_{n}$

*$b_1=-1,\space b_2=-5, ...,\space b_6=-29$

*$c_1=-4,\space c_2=16,\space c_3=-28,\space c_4=40,\space c_5=-52$
Once again, if you look at the sequence denoted by $c_n$, you can deduce the general term formula which is $c_n=(-1)^n(12n-8)$. Using this general term formula, we can easily write that $$\begin{align}\ b_{n+1}-b_{n}&= \ (a_{n+2}-a_{n+1})-(a_{n+1}-a_{n})\\
&=\ a_{n+2}-2a_{n+1}+a_{n}\\
&=\ (-1)^n(12n-8)\\
\end{align}$$ This last equation combined with our first deduction gives us $$4a_{n+1}=(-1)^n(8-12n)\Rightarrow a_{n+1}=(-1)^n(2-3n)$$ If we replace $n+1$ with $k$, we eventually get $$a_k=(-1)^{k}(3k-5)$$
