binomial theorem help? What's the 12th term? ?

for generic (x+y)^n 

kth term  = ( n  (k-1) )  x^((n- (k-1) )  y^((k-1) 

12th term 

n = 16 

k = 12 

k -1 = 11

(16 11)  x^ (16 -11)  y^11

(16 11) x^5  y^11  

16! / (11! 5!) =  16*15*14*13*12 / 5*4*3*2*1    

since 15  = 5*3*1  

and 16/(4*2)   = 2  

(16 11) =     2*14*13*12  =  4368

now substitute 

x^2 for x  

and 

(-1/2) for y 

4368 *( x^2) ^5  (-1/2)^11   

4368 *(x^10)   (-1)^11 / (2^11)   

(-4368/2048) *x^10  

Answer: 

-(273/128) x^10 

or 

(-2  17/128)x^10 

(x^2 - 1/2)^16

= x^32 - 8 x^30 + 30 x^28 - 70 x^26 + (455 x^24)/4 - (273 x^22)/2 + (1001 x^20)/8 - (715 x^18)/8 + (6435 x^16)/128 - (715 x^14)/32 + (1001 x^12)/128 - (273 x^10)/128 + (455 x^8)/1024 - (35 x^6)/512 + (15 x^4)/2048 - x^2/2048 + 1/65536

The 12th term is - (273 x^10)/128