Question

The minimum value of $ n $ for which the number of integer terms in the binomial expansion $\left(7^{\frac{1}{3}} + 11^{\frac{1}{12}}\right)^n$ is 183, is

• A. 2196
• B. 2172
• C. 2184
• D. 2148

Hint:

To find the number of integral terms in a binomial expansion with irrational components, use the formula for the terms and solve for \( n \).

Answer 1

The Correct Option is C

To find the smallest integer value of \( n \) such that the number of integer terms in the binomial expansion of \( \left(7^{\frac{1}{3}} + 11^{\frac{1}{12}}\right)^n \) is 183, we must follow these steps:

1. Consider the general term of the binomial expansion \( T_k = \binom{n}{k}(7^{\frac{1}{3}})^{n-k}(11^{\frac{1}{12}})^k \). For \( T_k \) to be an integer, both \( (7^{\frac{1}{3}})^{n-k} \) and \( (11^{\frac{1}{12}})^k \) must individually be integers.

2. This implies:

• \( \frac{n-k}{3} \) is an integer, i.e., \( n-k = 3m \) for some integer \( m \).
• \( \frac{k}{12} \) is an integer, i.e., \( k = 12p \) for some integer \( p \).

3. Solving these:

\( n - 3m = k = 12p \), thus \( n = 3m + 12p \).

4. To find the number of integer terms, \( k \) should vary such that both \( m \) and \( p \) are integers. Therefore, find all possible \( p \) satisfying \( 0 \leq 12p \leq n \).

5. The integer values of \( k = 12p \) must satisfy \( 0 \leq k \leq n \).

6. The condition is \( 0 \leq 12p \leq n \), resulting in \( 0 \leq p \leq \frac{n}{12} \).

7. Correspondingly, the condition \( n = 3m + 12p \) implies:

• The number of values of \( p \) is \( \left\lfloor \frac{n}{12} \right\rfloor + 1 \).

8. Given that the number of integer terms is 183, we equate:

\(\left\lfloor \frac{n}{12} \right\rfloor + 1 = 183 \)

9. Solve for \( n \):

\(\left\lfloor \frac{n}{12} \right\rfloor = 182 \)

This implies:

\(182 \times 12 \leq n < 183 \times 12\)

Resulting in:

\(2184 \leq n < 2196\)

10. Therefore, the smallest integer value for \( n \) is \( n = 2184 \).

The correct answer is: 2184.