JEE Mathematics 2023 Question with Solution

Given: We need the number of integral terms in the expansion of $\left( 3^{\frac{1}{2}} + 5^{\frac{1}{4}}\right)^{680}$.

Find: The count of terms that are integers.

The general term in the binomial expansion is

$$T_r = \binom{680}{r} \left( 3^{\frac{1}{2}} \right)^{680-r} \left( 5^{\frac{1}{4}} \right)^r$$

So,

$$T_r = \binom{680}{r} \cdot 3^{\frac{680-r}{2}} \cdot 5^{\frac{r}{4}}$$

For $T_r$ to be an integer, both exponents must be integers:

$$\frac{680-r}{2} \in \mathbb{Z}, \qquad \frac{r}{4} \in \mathbb{Z}$$

From $\frac{r}{4} \in \mathbb{Z}$, $r$ must be a multiple of $4$.

Also, if $r$ is a multiple of $4$, then $r$ is even, so $680-r$ is also even. Hence $\frac{680-r}{2}$ is automatically an integer.

Therefore,

$$r = 0, 4, 8, 12, \dots, 680$$

This is an arithmetic progression with first term $0$, last term $680$, and common difference $4$.

Number of values of $r$ is

$$\frac{680-0}{4} + 1 = 170 + 1 = 171$$

Therefore, the number of integral terms is $171$. The correct option is A.

Given: The expansion is $\left( 3^{\frac{1}{2}} + 5^{\frac{1}{4}}\right)^{680}$.

Find: How many terms are integral.

A term is integral only when the exponent of $5$ becomes an integer:

$$\frac{r}{4} \in \mathbb{Z}$$

So $r$ must be divisible by $4$.

Once $r$ is divisible by $4$, it is even, and hence $680-r$ is even. Therefore,

$$\frac{680-r}{2} \in \mathbb{Z}$$

also holds automatically.

So we only count multiples of $4$ from $0$ to $680$:

$$0, 4, 8, \dots, 680$$

Count:

$$\frac{680}{4} + 1 = 171$$

Therefore, the correct option is A.