Question

The number of integral terms in the expansion of \left( 3^{\frac{1}{2}} + 5^{\frac{1}{4}} \right)^{680} \text{ is equal to:}

Hint:

In binomial expansions, to find the number of integral terms, focus on the exponents of the terms, ensuring they yield integer results. For this, the exponents should be divisible by the corresponding denominators.

Answer 1

Correct Answer: 171

The general term in the binomial expansion of \( \left( 3^{\frac{1}{2}} + 5^{\frac{1}{4}} \right)^{680} \) is given by: \[ T_r = \binom{680}{r} \left( 3^{\frac{1}{2}} \right)^{680-r} \left( 5^{\frac{1}{4}} \right)^r \] Simplifying the exponents: \[ T_r = \binom{680}{r} \cdot 3^{\frac{680-r}{2}} \cdot 5^{\frac{r}{4}} \] For \( T_r \) to be an integer, both \( 3^{\frac{680-r}{2}} \) and \( 5^{\frac{r}{4}} \) must be integers. This means that \( \frac{680 - r}{2} \) and \( \frac{r}{4} \) must both be integers.
Thus, \( r \) must be a multiple of 4 for \( 5^{\frac{r}{4}} \) to be an integer. Additionally, \( 680 - r \) must be an even number for \( 3^{\frac{680-r}{2}} \) to be an integer.
Let \( r = 4k \), where \( k \) is an integer. We check for the values of \( r \) from 0 to 680 that satisfy this condition.
The values of \( r \) that satisfy \( r = 4k \) and \( r \leq 680 \) are \( 0, 4, 8, 12, \dots, 680 \).
Thus, the number of integral terms is given by the number of possible values of \( r \), which is \( 171 \).