JEE Mathematics 2024 Question with Solution

Given: We need the number of integral terms in the expansion of $\left( \sqrt{7}z + \frac{1}{6\sqrt{z}} \right)^{824}$.

Find: The number of terms for which the power of $z$ is an integer.

The general term is

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

Simplifying,

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

So,

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

For the term to be integral in $z$, the exponent

$$\frac{824-3k}{2}$$

must be an integer.

This requires $824-3k$ to be even. Since $824$ is even and $3k$ has the same parity as $k$, this means $k$ must be even.

Hence let

$$k = 2n$$

where $n = 0,1,2,\dots,412$.

Each such value gives one integral term. Therefore, the number of integral terms is

$$413$$

the solution states 138, but its working is inconsistent with the exponent obtained from the general term. Based on the extracted working expression, the count of integral terms should be $413$.

From

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

the exponent of $z$ is $\frac{824-3k}{2}$.

For this to be an integer, the numerator must be even:

$$824-3k \equiv 0 \pmod{2}$$

Since $824 \equiv 0 \pmod{2}$ and $3 \equiv 1 \pmod{2}$,

$$k \equiv 0 \pmod{2}$$

So only even values of $k$ are allowed.

The possible even values are

$$0,2,4,\dots,824$$

whose count is

$$\frac{824}{2}+1 = 413$$

Thus the number of integral terms is $413$.