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Expand using Binomial Theorem (1+x2-2x)4,x≠0

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Question

Expand using Binomial Theorem `(1+ x/2 - 2/x)^4, x != 0`

Sum
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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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Chapter 7: Binomial Theorem - Miscellaneous Exercise [Page 131]

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NCERT Mathematics [English] Class 11
Chapter 7 Binomial Theorem
Miscellaneous Exercise | Q 5. | Page 131

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