Question

If the number of terms in the expansion of \((x\sqrt{180} + \sqrt[3]{432})^{200}\) having integral coefficients is \(n\), then the value of \([n/6]\) is:

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

For integral coefficients, powers of square roots and cube roots must be integers. Combine conditions using LCM.

Answer 1

The Correct Option is B

Concept: General term in the expansion of \((a + b)^{200}\) is: \[ T_{r+1} = \binom{200}{r} a^{200-r} b^r \] Here \(a = x\sqrt{180}\) and \(b = \sqrt[3]{432}\).

Step 1: Simplify the roots. \[ \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \] \[ \sqrt[3]{432} = \sqrt[3]{216 \times 2} = 6\sqrt[3]{2} \]

Step 2: General term. \[ T_{r+1} = \binom{200}{r} (x \cdot 6\sqrt{5})^{200-r} (6\sqrt[3]{2})^r \] \[ = \binom{200}{r} x^{200-r} \cdot 6^{200-r} \cdot (\sqrt{5})^{200-r} \cdot 6^r \cdot (\sqrt[3]{2})^r \] \[ = \binom{200}{r} x^{200-r} \cdot 6^{200} \cdot 5^{\frac{200-r}{2}} \cdot 2^{\frac{r}{3}} \]

Step 3: Condition for integral coefficient. The coefficient is integral if:
•\(\frac{200-r}{2}\) is an integer \(\Rightarrow 200 - r\) is even \(\Rightarrow r\) is even.
•\(\frac{r}{3}\) is an integer \(\Rightarrow r\) is a multiple of 3. Also \(\binom{200}{r}\) is always an integer.

Step 4: Combine conditions. \(r\) must be a multiple of both 2 and 3 \(\Rightarrow\) multiple of 6. \[ r = 0, 6, 12, \dots, 198 \] Number of terms with integral coefficients: \[ n = \frac{198}{6} + 1 = 33 + 1 = 34 \]

Step 5: Compute \([n/6]\). \[ \left[\frac{34}{6}\right] = [5.666\ldots] = 5 \]