The value of greatest term (Numerically) in the expansion of (3+2x)80 when x=1

(3+2x)80=(3)801+2x380, here  n=80;X=2x3 When x=1;X=23 Now (n+1)|X||X|+1=(80+1)×2323+1=81·2353=1625=32·4=32+0.4=p+F∴ The greatest term is Tp+1=T32+1=T33T33=T32+1      =C32   80380−32.(2x)32      =C32   80348.(2)32x32 At x=1,T33=C32   80348·(2)32