Answer 1

Let t_r+1 contains x^–3 in the expansion of `(x - 1/(2x))^5`.

We know that, in the expansion of (a + b)ⁿ,

t_r+1 = ⁿC_r a^n–r b^r

Here a = x, b = `-1/(2x)`, n = 5

∴ t_r+1 = `""^5"C"_"r" (x)^(5 - "r") (-1/(2x))^"r"`

= `""^5"C"_"r"  x^(5 - "r").(-1/2)^"r".x^(-"r")`

= `""^5"C"_"r".(-1/2)^"r".x^(5 - 2"r")`

But t_r+1 contains x^–3

∴ power of x = – 3

∴ 5 – 2r = – 3

∴ r = 4

∴ coefficient of x^–3 = `""^5"C"_4.(-1/2)^4`

= `5 xx 1/16`

= `5/16`