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Question
Expand using Binomial Theorem `(1+ x/2 - 2/x)^4, x != 0`
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Solution
`(1 + x/2 - 2/x)^4 = [(1 + x/2) - 2/x]^4`
= `(1 + x/4)^4 + ^4C_1 (1 + x/2)^3 (-2/x) + ^4C_2 (1 + x/2)^2 (-2/x)^2 + ^4C_3 (1 + x/2) (-2/x)^3 + ^4C_4 (-2/x)^4`
= `(1 + x/2)^4 + 4(1 + x/2)^3 (-2/x) + 6 (1 + x/2)^2 (4/x^2) + 4 (1 + x/2) (- 8/x^3) + (16/x^4)`
= `(1 + x/2)^4 , (1 + x/2)^3 , (1 + x/2)^2` on spreading
= `(1 + x/2 - 2/x)^4 = (1 + 4. x/2 + 6 x^2/4 + 4. x^3/8 + x^4/16) - 8/x (1 +3 . x/2 + 3. x^2/4 + x^3/8) + 24/x^2 (1 + x + x^2/4) - 32/x^3 (1 + x/2) + 16/x^4`
= `(1 + 2x + 3/2 x^2 + 1/2 x^3 + x^4/16) - 8/x(1 + 3/2 x + 3/4 x^2 + x^3/8) + 24/x^2 (1 + x + x^2/4) - 32/x^3 (1 + x/2) + 16/x^4`
= `(1 + 2x + 3/2 x^2 + 1/2 x^3 + x^4/16) - (8/x + 12 + 6x + x^2) + (24/x^2 + 24/x + 6) - (32/x^3 + 16/x^2) + 16/x^4`
= `x^4/16 + x^3/2 + (3/2 - 1)x^2 + (2 -6)x + (1 - 12 +6) + (- 8 + 24) 1/x + (24 -16) 1/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`
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