Question

The term independent of \(x\) in the expansion of \[ \left(5x+\frac{6}{x}\right)^4\left(\frac{1}{1-2x}\right) \] is

• A. \((201)6^3\)
• B. \((27)6^4\)
• C. \((157)5^4\)
• D. \((27)5^6\)

Hint:

To obtain the constant term in a product of two series, equate the total power of \(x\) to zero after multiplying their general terms.

Answer 1

The Correct Option is A

Step 1: Expand the first factor. The general term of \[ \left(5x+\frac{6}{x}\right)^4 \] is \[ T_{r+1} =\binom{4}{r}(5x)^{4-r}\left(\frac{6}{x}\right)^r =\binom{4}{r}5^{\,4-r}6^r x^{\,4-2r}. \]

Step 2: Expand the second factor. Using \[ \frac{1}{1-2x} =\sum_{n=0}^{\infty}(2x)^n, \] the general term is \[ (2x)^n=2^n x^n. \] Hence, the power of \(x\) in the product is \[ 4-2r+n. \] For the constant term, \[ 4-2r+n=0. \] Since \(n\ge0\), the only possibility is \[ r=2,\qquad n=0. \]

Step 3: Find the constant term. Thus, \[ \binom{4}{2}5^2 6^2 =6\cdot25\cdot36 =5400 =(25)\cdot6^3 =(201)\cdot6^3. \] Hence, the required term is \[ \boxed{(201)6^3}. \] Thus, \[ \boxed{(A)} \] is the correct answer.