Using binomial theorem expand `(1+x/2-2/x)^4,\ x!=0.`

Explanation

We know that

`(a+b)^n=^nC_0 a^n+^nC_1 a^(n-1)b^1+^nC_2 a^(n-2)b^2+ .............^nC_(n-1) a^1b^(n-1)+^nC_(n)b^(n)`

Hence,

`(a+b)^4=^4C_0 a^4+^4C_1 a^(3)b^1+^4C_2 a^(2)b^2+^4C_(3) a^1b^(3)+^4C_(4)b^(4)`

` " " " " ``= (4!)/(0!(4-0)!)a^4+(4!)/(1!(4-1)!)a^3 b^1+(4!)/(2!(4-2)!)a^2b^2+(4!)/(3!(4-3)!)a^1b^3+(4!)/(4!(4-4)!)b^4`

` " " " " ``= (4!)/(1xx4!)a^4+(4!)/(1xx3!)a^3 b^1+(4!)/(2!2!)a^2b^2+(4!)/(3!1!)a^1b^3+(4!)/(4!0!)b^4`

` " " " " ``= a^4+4a^3 b^1+6a^2b^2+4a^1b^3+b^4 " " " " .....(1)`

Hence,

`(a+b)^4= a^4+4a^3 b^1+6a^2b^2+4a^1b^3+b^4`

We need to find `((1+x/2)-2/x)^4`

Putting `a=(1+(x)/2)` and `b=((-2)/x)`


`((1+(x)/2)+((-2)/x))^4`

` " " " " ``=(1+x/2)^4+4(1+x/2)^3((-2)/x)+6(1+x/2)^2((-2)/x)^2+4(1+x/2)^1((-2)/x)^3+((-2)/x)^4`

` " " " " ` `=(1+x/2)^4-8/x(1+x/2)^3+(24)/x^2(1+x/2)^2-(32)/x^3(1+x/2)+(16)/x^4`


Now Solving `(1+x/2)^4, (1+x/2)^3` separately


Solving `(1+x/2)^4`

From (1), `(a+b)^4= a^4+4a^3 b^1+6a^2b^2+4a^1b^3+b^4`

Putting `a=1` and `b=x/2`

`(1+x/2)^4=(1)^4+4(1)^3(x/2)+6(1)^2(x/2)^2+4(1)^1(x/2)^3+(x/2)^4`

` " " " " ``=1+4(x/2)+6(x^2/4)+4(x^3/8)+(x^4/16)`

` " " " " ` `=1+2x+3/2x^2+x^3/2+x^4/16`


Now Solving `(1+x/2)^3`

From (1), `(a+b)^3= a^3+3a^2 b+3ab^2+b^3`

Putting `a=1` and `b=x/2`

`(1+x/2)^3= (1)^3+3(1)^2 (x/2)+3(1)(x/2)^2+(x/2)^3`

` " " " " ``=1+(3x)/2+(3x^2)/4+(x^3)/8`


Now,

`((1+x/2)-2/x)^4=(1+x/2)^4-8/x(1+x/2)^3+(24)/x^2(1+x/2)^2-(32)/x^3(1+x/2)+(16)/x^4`

Putting value of `(1+x/2)^4, (1+x/2)^3`

` " " " " ``=(1+2x+3/2x^2+x^3/2+x^4/16)-(8/x+12+6x+x^2)+((24)/x^2+(24)/x^2xx x^2/4+(24)/x^2xx x)-((32)/x^3+16/x^2)+(16)/x^4`

` " " " " ``=1+2x+3/2x^2+x^3/2+x^4/16-8/x-12-6x-x^2+(24)/x^2+6+(24)/x-(32)/x^3-16/x^2+(16)/x^4`

` " " " " ` `=x^4/16+x^3/2+3/2x^2-x^2+2x-6x+1-12+6-8/x+(24)/x+(24)/x^2-16/x^2-(32)/x^3+(16)/x^4`

` " " " " ` `=x^4/16+x^3/2+x^2/2-4x-5+(16)/x+(8)/x^2-(32)/x^3+(16)/x^4`


Thus,

`((1+x/2)-2/x)^4=x^4/16+x^3/2+x^2/2-4x-5+(16)/x+(8)/x^2-(32)/x^3+(16)/x^4`