Using Binomial Theorem, the given expression (1+2x−x2)4 can be expanded as
[(1+2x)−x2]4
= 4C0(1+2x)4− 4C1(1+2x)3(x2)+ 4C2(1+2x)2(x2)2− 4C3(1+2x)(x2)3+ 4C4(x2)4
=(1+2x)4−4(1+2x)3(x2)+6(1+x+4x2)(x24)−4(1+2x)(x38)+x416
= (1+2x)4 −x8((1+2x)3+x224+x24+6−x332−x216+x416
= (1+2x)4 −x8((1+2x)3+x28+x24+6+x332+x416 ......(1)
Again by using Binomial Theorem, we obtain
(1+2x)4= 3C0(1)4+ 4C1(1)3(2x)+ 4C2(1)2(2x)2 + 4C3(1)(2x)3 + 4C4(2x)4 =1+4×2x+6×4x2+4×8x3+16x4 ...(2)
(1+2x)3= 3C0(1+2x)3− 3C1(1)2(2x)+ 3C2(1)(2x)2− 3C3(2x)3 =1+23x+43x2+8x3 ....(3)
from (1), (2) and (3), we obtain
[(1+2x)−x2]4
=1+2x+23x2+2x3+16x4−x8(1+23x+43x2+8x3)+x28+x24+6−x332+x416
= 1+2x+23x2+2x3+16x4−x8−12−6x−x2+x28+x24+6−x332+x416
=x16+x28−x332+x416−4x=2x2+2x3+16x4−5