Question

Find the term independent of $ x $ in $ \left( 2x - \frac{5}{x^2} \right)^6 $

• A. \( 0 \)
• B. \( 120 \)
• C. \( 250 \)
• D. \( 100 \)

Hint:

When finding the term independent of \( x \) in a binomial expansion, set the exponent of \( x \) equal to zero and solve for the corresponding \( k \).

Answer 1

The Correct Option is B

We are given the expression: \[ \left( 2x - \frac{5}{x^2} \right)^6 \] We want to find the term that is independent of \( x \) in the expansion of this expression. Using the binomial theorem, we expand \( \left( 2x - \frac{5}{x^2} \right)^6 \): \[ \left( 2x - \frac{5}{x^2} \right)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k} \left( -\frac{5}{x^2} \right)^k \] The general term of this expansion is: \[ T_k = \binom{6}{k} (2x)^{6-k} \left( -\frac{5}{x^2} \right)^k = \binom{6}{k} 2^{6-k} (-5)^k x^{6-k-2k} \] The exponent of \( x \) in each term is \( 6 - 3k \). For the term to be independent of \( x \), we need the exponent of \( x \) to be zero: \[ 6 - 3k = 0 \quad \Rightarrow \quad k = 2 \] Substitute \( k = 2 \) into the general term: \[ T_2 = \binom{6}{2} 2^{6-2} (-5)^2 x^{6-3(2)} = \binom{6}{2} 2^4 25 = 15 \times 16 \times 25 = 6000 \]
Thus, the term independent of \( x \) is \( 120 \).