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 (x²/4) + 4 (x³/8) + (x⁴/16)
  = 1+ 2x + 3x²/2 + x³/2 + x⁴/16
   
• (1 + x/2)³ = 1³ + 3(1)² (x/2) + 3 (1) (x/2)² + (x/2)³ (Using (a+b)³ formula)
  = 1 + 3x/2 + 3x²/4 + x³/8
   
• (1+ x/2)² = 1² + 2 (1)(x/2) + (x/2)² (Using (a+b)² formula)
  = 1 + x + x²/4

Substitute each of these in (1),

[(1 + x/2) - 2/x ]⁴ 

= [1+ 2x + 3x²/2 + x³/2 + x⁴/16] + 4 [1 + 3x/2 + 3x²/4 + x³/8] (-2/x)

+ 6 [1 + x + x²/4] (4/x²) + 4 (1 + x/2) (-8/x³) + 16/x⁴

= [1+ 2x + 3x²/2 + x³/2 + x⁴/16] + [-8/x - 12 - 6x - x²]

+ [24/x² + 24/x + 6] + [-16/x² - 32/x³] + 16/x⁴

= 16/x + 8/x² - 32/x³ + 16/x⁴ - 4x + x²/2 + x³/2 + x⁴/16 - 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/x² - 32/x³ + 16/x⁴ - 4x + x²/2 + x³/2 + x⁴/16 - 5