find the coeffcient of x^2 in the expansion of (1+x^2)(x/2 - 4/x)^6?

Method 1:

(1 + x²)[(x/2) - (4/x)]⁶
= (1 + x²)[(x/2)⁶ - ₆C₁ (x/2)⁵ (4/x) + ₆C₂ (x/2)⁴ (4/x)² - ₆C₃ (x/2)³ (4/x)³ + ₆C₄ (x/2)² (4/x)⁴ - ₆C₅ (x/2) (4/x)⁵ + (4/x)⁶]
= (1 + x²)[…… + ₆C₂ (x/2)⁴ (4/x)² - ₆C₃ (x/2)³ (4/x)³ + ……]
= (1 + x²)[…… + 15 (x⁴/16) (16/x²) - 20 (x³/8) (64/x³) + ……]
= (1 + x²)[…… + 15x² - 160 + ……]

The coefficient of x² in the expansion of (1 + x²)[(x/2) - (4/x)]⁶
= 15 - 160
= -145

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Method 2:

The nth term of [(x/2) - (4/x)]⁶ = (-1)ⁿ⁺¹ ₆Cₙ₋₁ (x/2)⁶⁻ⁿ⁺¹ (4/x)ⁿ⁻¹

When [(x/2) - (4/x)]⁶ = x⁰
6 - n + 1 - (n - 1) = 0
n = 4
The constant term of [(x/2) - (4/x)]⁶ = (-1)⁵ ₆C₃ (x/2)³ (4/x)³ = -160

When [(x/2) - (4/x)]⁶x⁶⁻ⁿ⁺¹ (1/x)ⁿ⁻¹ = x²
6 - n + 1 - (n - 1) = 2
n = 3
The x² term of [(x/2) - (4/x)]⁶= (-1)⁴ ₆C₂ (x/2)⁴ (4/x)² = 15

(1 + x²)[(x/2) - (4/x)]⁶
= (1 + x²)(-160 + 15x² + …… )

The coefficient of x² in the expansion of (1 + x²)[(x/2) - (4/x)]⁶
= 15 - 160
= -145

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