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