here's the sequence/pattern:
1, -1, 2, -2, 3,....
how would u explain the pattern besides the fact that 
it goes in order and after the positive number you put 
the opposite of it?? or is that all?? or is some 
addition/subtraction involved??

Let's find the general term for an

a1 = 1, a3 = 2, a5 = 3, and if n is odd, an = (n+1)/2

a2 = -1, a4 = -2, a6 = -3, and if n even, an = -n/2

So when n is odd, we want 1 times the formula (n+1)/2 
plus 0 times the formula -n/2,

and when n is even, we want 0 times the formula (n+1)/2 
plus 1 times the formula -n/2.

To do that, we make use of 2 special sequences 
that alternate 0's and 1's:

1, 0, 1, 0, 1, 0, 1, ...

which has general term [1 + (-1)n+1]/2

and this sequence 

0, 1, 0, 1, 0, 1, 0, ...

which has general term [1 + (-1)n]/2

We showed above that

if n is odd, an = (n+1)/2

and if n is even, an = -n/2

So when n is odd, we want 1 times 
the "odd" formula plus 0 times the 
"even" formula,

and 

when n is even, we want 1 times 
the "even" formula plus 0 times the 
odd "formula".

We can do that by multiplying
the general term of the 1,0,1,0,...
sequence by the odd formula
and the general term of the 
0,1,0,1,... sequence by the even
formula, and adding them


  1+(-1)n+1   n+1     1 + (-1)n   -n 
-----------·----- + -----------· --- 
     2        2         2         2 

Multiplying numerators and denominators,
use FOIL on first and distribute on
second

 n + 1 + (-1)n+1n + (-1)n+1     -n - (-1)nn
--------------------------- + ------------- 
              4                     4

Combine all over the LCD of 4


 n + 1 + (-1)n+1n + (-1)n+1  - n - (-1)nn
------------------------------------------ 
                     4                     

The n and -n cancel

 1 + (-1)n+1n + (-1)n+1 - (-1)nn
---------------------------------
               4

The last term of the numerator -(-1)nn can
be written as +(-1)n+1n

 1 + (-1)n+1n + (-1)n+1 + (-1)n+1n
------------------------------------
                4 

 1 + 2(-1)n+1n + (-1)n+1 
-------------------------
             4

Factoring (-1)n+1 out of the last
two terms.

      1 + (-1)n+1(2n + 1) 
an = ---------------------
              4

That's your formula for an, the nth term

Edwin