We know that the (r + 1)th term in the expansion of (x+a)n is given by
Tr+1 =

n

‌

Crxn−rar
Then (r + 1)th term in the expansion of (x+(−2y))12 is given by
Tr+1 =

12

‌

Crx12−r(−2y)r
For 4th term, we have r + 1 = 4 ⇒ r = 3
Substituting r = 3 in (i), we get
T4 =

12

‌

C3x12−3(−2y)3 = 220x9(−8)y3 = - 1760 x9y3