Question

The sum of the coefficients of all even degree terms in x in the expansion of

\({\left( {{\rm{x}} + \sqrt {{{\rm{x}}^3} - 1} } \right)^6} + {\left( {{\rm{x}} - \sqrt {{{\rm{x}}^3} - 1} } \right)^6}\), (x > 1) is equal to:

1. 29
2. 32
3. 26
4. 24

Answer 1

Correct Answer - Option 4 : 24

Binomial expression is an algebraic expression consisting of two different terms. “Different terms” here means either the two terms should have different variables or the different powers on its variable.

We have to find the sum of the coefficient of even parts.

On expanding the equation,

\({\left( {{\rm{x}} + \sqrt {{{\rm{x}}^3} - 1} } \right)^6} - {\left( {{\rm{x}} + \sqrt {{{\rm{x}}^3} - 1} } \right)^6}\) 

\(\Rightarrow ({{\rm{\;}}^6}{{\rm{C}}_0}\cdot{{\rm{x}}^6} + {{\rm{\;}}^6}{{\rm{C}}_1}\cdot{{\rm{x}}^5}\sqrt {{{\rm{x}}^3} - 1{\rm{\;}}} \ldots ) + ({{\rm{\;}}^6}{{\rm{C}}_0}\cdot{{\rm{x}}^6} - {{\rm{\;}}^6}{{\rm{C}}_1}\cdot{{\rm{x}}^5}\sqrt {{{\rm{x}}^3} - 1{\rm{\;}}} \ldots )\) 

After cancelling the odd terms, the only term that will remain is:

2(⁶C₀ x⁶ + ⁶C₂ x⁴(x³ – 1) + ⁶C₄ x²(x³ – 1)² + ⁶C₆ (x³ – 1)³)      ----(1)

Now, we have to pick the even powers of x

We know that, (x³ – 1)² = x⁶ – 2x³ + 1

Equation (1) becomes,

x = 2(⁶C₀ x⁶ – ⁶C₂ x⁴ + ⁶C₄ x⁸ + ⁶C₂ x² + ⁶C₆ (-1 – 3x⁶)

= 2(⁶C₀ x⁶ - ⁶C₂ x⁴ + ⁶C₄ x⁸ + ⁶C₂ x² – ⁶C₆ – 3 ⁶C₆ x⁶

When we drop the x terms and add the coefficient,

2(1 – ⁶C₂ + ⁶C₄ + ⁶C ₄ – 1 – 3)

= 2(1 – 15 + 15 + 15 – 1 – 3)

= 2(1 + 15 – 4)

= 2(12) = 24