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

Step 1: Understanding the Concept:
A term in a binomial expansion \((a + b)^{N}\) has an integral coefficient if the powers of the irrational parts of \(a\) and \(b\) result in rational numbers.
Step 2: Detailed Explanation:
The general term is \(T_{r+1} = \binom{200}{r} (x\sqrt{180})^{200-r} (\sqrt[3]{432})^{r}\).
For the coefficient to be an integer, the irrational parts must become rational.
1. \(\sqrt{180} = (180)^{1/2}\). For this to be rational, its power \((200-r)\) must be even. This implies \(r\) must be even.
2. \(\sqrt[3]{432} = (432)^{1/3}\). For this to be rational, its power \(r\) must be a multiple of 3.
3. Combining both: \(r\) must be a multiple of 2 AND a multiple of 3 \(\implies r\) is a multiple of 6.
Values of \(r \in \{0, 6, 12, \dots, 198\}\).
This is an AP: \(198 = 0 + (n-1)6 \implies 33 = n-1 \implies n = 34\).
Now find \([n/6]\):
\[ [34/6] = [5.666\dots] = 5 \]
Step 3: Final Answer:
The value is 5.