[1 + (x/2) – (2/x)]4
= [1 + {(x/2) – (2/x)}]4
= 4c0 + 4c1 [(x/2) – (2/x)] + 4c2 [(x/2) – (2/x)]2 + 4c3 [(x/2) – (2/x)]3(2/x)]3 + 4c4 [(x/2) – (2/x)]4
= 1 + 4[(x2 – 4) / (2x)] + 6[(x4 – 8x2 + 16) / (4x2)] + [4 / (8x3)] (x6 – 12x4 + 48x2 – 64) + (x4 / 16) – x2 + 6 – (16 / x2) + (16 / x4)
= 1 + 2x – (8/x) + (3/2)x2 – 12 + (24 / x2) + (x3 / 2) – 6x + (24 / x) – (32 / x3) + (x4 / 16) – x2 + 6 – (16 / x2) + (16 / x4)
= – 5 – 4x + (16 / x) + (x2 / 2) + (x3 / 2) + (x4 / 16) + (8 / x2) – (32/ x3) + (16 / x4)
Hence constant term is – 5.