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

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

Simplifying the expression for the general term, we have:

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

This can be expressed as:

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

For \( T_k \) to be an integer, the exponent of \( z \), \( (824-3k)/2 \), must be an integer. This implies \( 824-3k \) must be even. Simplifying, \( 824-3k = 2m \) for some integer \( m \), giving:

\( 3k = 824-2m \)

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

\( k \) is an integer, so \( 824-2m \) must be divisible by 3. Solving for \( m \) from \( 3k = 824-2m \) and simplifying, \( m \equiv 412 \pmod{3} \) or \( m \equiv 1 \pmod{3} \). Hence, \( m = 3n+1 \) for integer \( n \). Now:

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

\( 822-6n = 3k \)

\( k = 274-2n \)

\( k \) ranges from 0 to 824, giving valid \( n \) between 0 and 137. Each \( n \) gives a unique integral term.

Thus, the number of integral terms is: 138.

Answer 2

The general term in the expansion is:

\[ t_{r+1} = \binom{824}{r} \left( \sqrt{7} z \right)^{824 - r} \left( \frac{1}{6 \sqrt{z}} \right)^r \]

which simplifies to \( z^{\frac{824 - r}{7} - \frac{r}{6}} \).

For an integral power, \( r \) must be a multiple of 6.

Thus, \( r = 0, 6, 12, \dots, 822 \), giving 138 integral terms.