$n$-th term of an infinite sequence Determine the $n$-th term of the sequence $1/2,1/12,1/30,1/56,1/90,\ldots$. I have not been able to find the explicit formula for the $n$-th term of this infinite sequence. Can some one solve this problem and tell me how they find the answer?
 A: I would bet on the inverses of OEIS A002939, so $a_n = \frac{1}{2n(2n-1)}$  I just typed $2,12,30,56,90$ into the search box.  There were three hits.
A: To find a formula you need to first find the pattern.
There are a number of ideas to use when trying to find for such paterns in a sequence:


*

*if all numbers are composite, try to factorise - if this works nicely a pattern may emerge at this stage, especially for numbers with one prime factor, e.g. $56 = 7 \times 8$ 

*often a sequence will be built on the natural numbers $1,2,3,\dots$ somehow

*use trial and error, and look at the context around the numbers for any clues


Using these ideas, you could come up with 


*

*$2 = 1 \times 2$

*$12 = 3 \times 4$

*$30 = 5 \times 6$

*$56 = 7 \times 8$ 


If you want to pair these with the natural numbers to develop a formula, you will need to recognise that you are dealing with pairs of consecutive numbers that ascend by two from the previous pair.
$n: 1,2,3,4,5,\dots \\
2n: 2,4,6,8,10,\dots \\$
Hence, each pair is $\{2n-1,2n\}$.
Then $s_n = \dfrac{1}{2n-1} \cdot \dfrac{1}{2n}$  
