Question

Which term in the expansion of \(( x - \frac{1}{x^3})^{40}\) is independent of \(x\)?

• A. 7th
• B. 8th
• C. 9th
• D. 11th

Hint:

For binomial expansions, find the term where the exponent of \( x \) is zero to determine the term independent of \( x \).

Answer 1

The Correct Option is D

We need to find the term that is independent of \( x \) in the expansion of \( \left( x - \frac{1}{x^3} \right)^{40} \). Using the binomial expansion, the general term is given by: \[ T_{r+1} = \binom{40}{r} x^{40 - r} \left( - \frac{1}{x^3} \right)^r. \] Simplifying: \[ T_{r+1} = \binom{40}{r} x^{40 - r} \times (-1)^r \times x^{-3r} = \binom{40}{r} (-1)^r x^{40 - r - 3r}. \] The exponent of \( x \) is \( 40 - 4r \). For the term to be independent of \( x \), we set the exponent of \( x \) equal to zero: \[ 40 - 4r = 0 \quad \Rightarrow \quad r = 10. \] Thus, the term that is independent of \( x \) is the \( (r+1) \)-th term, which is the 11th term.