# $n^{th}$ term for the following sequence

I'm trying to find a formula for the $n^{th}$ term of following sequence:

$$\frac{-1}{2}, 0, \frac{1}{10}, 0, \frac{-1}{26}, 0, \frac{1}{50}, 0, \frac{-1}{82}, 0, \frac{1}{122}, 0, \dots$$

It looks like every other term the uses the pattern

$1^2+1, 3^2+1, 5^2+1, 7^2+1 \dots$ in the denominator.

I'm not sure how to go about putting this all together as a sequence. It has to be something simple that I'm just not thinking of at the moment. Any help would be appreciated. Thanks.

• Why must you put the sequence together with a formula? The words describing every other denominator tell your reader exactly what you want. The formulas proposed in the various correct answers are hard for readers to figure out and would detract from whatever you are trying to say. (There are many questions on this site asking for similar formulas, for bad reasons.) – Ethan Bolker Nov 1 '17 at 18:32
• While I agree with most of what you're saying, I was hoping to see a formula since this is how it was asked in a lecture. I don't think that's too much to ask. – metausar Nov 1 '17 at 18:54

## 2 Answers

You found the important part. So your sequence is $1/(n^2+1)$, except that is multiplied with $\pm1$ or $0$. This is something like $-\sin(n\pi/2)$ So your sequence is $$-\frac{\sin (n\pi/2)}{n^2+1}$$

$$\frac{(-1)^{(n+1)/2}}{1+n^2}$$ for the odd terms. Zero for the even terms.