# 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 – user161695 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 – user161695 Jul 30 '15 at 23:55

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 $$a_k=(-1)^{k}(3k-5)$$