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

**Solution:**

By using the binomial theorem, the expansion of the given expression is,

[(1 + x/2) - 2/x ]^{4} = ⁴C₀ (1 + x/2)^{4} + ⁴C₁ (1 + x/2)^{3} (-2/x) + ⁴C₂ (1 + x/2)^{2} (-2/x)^{2} + ⁴C₃ (1 + x/2) (-2/x)^{3} + ⁴C₄ (-2/x)^{4}

Using the nCr formula,

[(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) (-2/x)^{3} + (-2/x)^{4} ... (1)

Now, we will expand (1 + x/2)^{4} again using the binomial theorem. Then,

- (1 + x/2)
^{4}= ⁴C₀ (1)^{4}+ ⁴C₁ (1)^{3}(x/2) + ⁴C₂ (1)^{2}(x/2)^{2}+ ⁴C₃ (1) (x/2)^{3}+ ⁴C₄ (x/2)^{4}

= 1 + 4 (x/2) + 6 (x^{2}/4) + 4 (x^{3}/8) + (x^{4}/16)

= 1+ 2x + 3x^{2}/2 + x^{3}/2 + x^{4}/16

- (1 + x/2)
^{3}= 1^{3}+ 3(1)^{2}(x/2) + 3 (1) (x/2)^{2}+ (x/2)^{3}(Using (a+b)³ formula)

= 1 + 3x/2 + 3x^{2}/4 + x^{3}/8

- (1+ x/2)
^{2}= 1^{2}+ 2 (1)(x/2) + (x/2)^{2}(Using (a+b)² formula)

= 1 + x + x^{2}/4

Substitute each of these in (1),

[(1 + x/2) - 2/x ]^{4}

= [1+ 2x + 3x^{2}/2 + x^{3}/2 + x^{4}/16] + 4 [1 + 3x/2 + 3x^{2}/4 + x^{3}/8] (-2/x)

+ 6 [1 + x + x^{2}/4] (4/x^{2}) + 4 (1 + x/2) (-8/x^{3}) + 16/x^{4}

= [1+ 2x + 3x^{2}/2 + x^{3}/2 + x^{4}/16] + [-8/x - 12 - 6x - x^{2}]

+ [24/x^{2} + 24/x + 6] + [-16/x^{2 }- 32/x^{3}] + 16/x^{4}

= 16/x + 8/x^{2 }- 32/x^{3} + 16/x^{4 }- 4x + x^{2}/2 + x^{3}/2 + x^{4}/16 - 5

NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 9

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

**Summary:**

We expanded and found the value of (1 + x/2 - 2/x)⁴ to be 16/x + 8/x^{2 }- 32/x^{3} + 16/x^{4 }- 4x + x^{2}/2 + x^{3}/2 + x^{4}/16 - 5

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