The value of greatest term(Numerically) in the expansion of 21+1210

We have n=10,x=12 Now (n+1)|x||x|+1=(10+1)×1212+1=11×121+22=112.414=4.55=4+0.55=p+F∴ Numerically greatest term is Tp+1=T5 Now T5=T4+1=210C4126=2·C4   10(2)6=C4   10(2)5=21042=10524