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 determine the minimum integer value of \( n \) for which the binomial expansion of \( \left(7^{\frac{1}{3}} + 11^{\frac{1}{12}}\right)^n \) contains 183 integer terms, the following procedure is followed:

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

2. This necessitates the following conditions:

• \( \frac{n-k}{3} \) must be an integer, implying \( n-k = 3m \) for some integer \( m \).
• \( \frac{k}{12} \) must be an integer, implying \( k = 12p \) for some integer \( p \).

3. Solving these two equations yields:

\( n - 3m = k = 12p \), which rearranges to \( n = 3m + 12p \).

4. To count the integer terms, \( k \) must vary such that \( m \) and \( p \) are integers. This means we need to find all possible integer values of \( p \) such that \( 0 \leq 12p \leq n \).

5. The integer values of \( k \) must be within the range \( 0 \leq k \leq n \).

6. The condition \( 0 \leq 12p \leq n \) implies \( 0 \leq p \leq \frac{n}{12} \).

7. Considering the condition \( n = 3m + 12p \), the number of possible values for \( p \) is determined by:

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

8. Given that the total count of integer terms is 183, we set up the equation:

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

9. Solving for \( n \):

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

This inequality holds:

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

Which simplifies to:

\(2184 \leq n < 2196\)

10. Consequently, the smallest integer value of \( n \) that satisfies the condition is \( n = 2184 \).

The determined answer is: 2184.