Question

Number of integral terms in the expansion of \( \left( \sqrt{7} z + \frac{1}{6 \sqrt{z}} \right)^{824} \) is equal to ______.

Answer 1

Correct Answer: 138

To determine the count of integral terms in the expansion of \( \left( \sqrt{7} z + \frac{1}{6 \sqrt{z}} \right)^{824} \), we examine the general term of the binomial expansion:

\( T_k = \binom{824}{k} (\sqrt{7}z)^{824-k} \left( \frac{1}{6\sqrt{z}} \right)^k \)

Simplifying this expression yields:

\( T_k = \binom{824}{k} (7^{(824-k)/2}) z^{(824-k)-k/2} \frac{1}{6^k} \)

Which can be rewritten as:

\( T_k = \binom{824}{k} \frac{7^{(824-k)/2}}{6^k} z^{(824-3k)/2} \)

For \( T_k \) to be integral, the exponent of \( z \), \( (824-3k)/2 \), must be an integer. This requires \( 824-3k \) to be an even number. Let \( 824-3k = 2m \) for some integer \( m \). This leads to:

\( 3k = 824-2m \)

\( k = \frac{824-2m}{3} \)

Since \( k \) must be an integer, \( 824-2m \) must be divisible by 3. Solving \( 3k = 824-2m \) for \( m \) gives \( m \equiv 412 \pmod{3} \), which simplifies to \( m \equiv 1 \pmod{3} \). Therefore, \( m \) can be expressed as \( m = 3n+1 \) for an integer \( n \). Substituting this back into the equation for \( k \):

\( 824-2(3n+1) = 3k \)

\( 822-6n = 3k \)

\( k = 274-2n \)

Given that \( k \) ranges from 0 to 824, the valid range for \( n \) is from 0 to 137. Each distinct value of \( n \) corresponds to a unique integral term.

Consequently, the total number of integral terms is: 138.