Using binomial theorem, expand each of the following:(1+x/2-2/x)^(4),x

1.	Using binomial theorem, expand each of the following:(1+x/2-2/x)^(4),x!=0
Answer» Solution :`(1+x/2-2/x)^(4) = {1 + (x/2-2/x)}^(4)` `(1+y)^(4) " where " (x/2 - 2/x)=y` `=^(4) C_(0) + .^(4)C_(1)y + .^(4)C_(2)y^(2) + .^(4) C _(3)y^(3) + .^(4)C_(4)y^(4)` `1+ 4y+ 6y^(2) + 4y^(3) + y^(4)` ` 1+ 4 (x/2-2/x)+ 6 ( x/2-2/x)^(2) + 4(x/2- 2/x)^(3) + ( x/2-2/x)^(4)` `=1+ ( 2x - 8/x) + 6 ( x^(2)/4 + 4/x^(2)-2) + 4[.^(3)C_(0)(x/2)^(3)-.^(3)C_(1) (x/2)^(2) (2/x)+.^(3)C_(2)(x/2)(2/x)^(2) -.^(3) C_(3) (2/x)^(3)]` `[.^(4) C_(0) (x/2)^(4)-.^(4)C_(1)(x/2)^(3) (2/x)+.^(4)C_(2)(x/2)^(2)(2/x)^(2)-.^(4)C_(3) (x/2)(2/x)^(3) + .^(4)C_(4)(2/x)^(4)]` `= 1+(2x-8/x)+((3X^(2))/2+ 24/x^(2) - 12 ) + 4 [ x^(3)/8 - 3/2 x +6.x - 8/x^(3)] + [x^(4)/16 - x^(2) + 6 - 16/x^(2) + 16/x^(4)]` `=1 + (2x-8/x)+ ((3x^(2))/2 +24/x^(2) - 12)+(x^(3)/2-6x+24/x-32/x^(3)) +(x^(4)/16 - x^(2) + 6 - 16/x^(2) + 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)`

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