Question

The number of rational terms in the expansion of $(\sqrt{3}+\sqrt[4]{5})^{124}$ is

• A. 31
• B. 32
• C. 33
• D. 34

Answer 1

Answer: (B)

32

Answer 2

The general term in the expansion of $(\sqrt{3}+\sqrt[4]{5})^{124}$ is given by: $T_{r+1} = \binom{124}{r} (\sqrt{3})^{124-r} (\sqrt[4]{5})^r = \binom{124}{r} 3^{\frac{124-r}{2}} 5^{\frac{r}{4}}$ For the term to be rational, the exponents of 3 and 5 must be integers. This means $\frac{124-r}{2}$ must be an integer, so $124-r$ must be even. Since 124 is even, $r$ must be even. Also, $\frac{r}{4}$ must be an integer, so $r$ must be a multiple of 4. If $r$ is a multiple of 4, it is automatically even. Thus, the only condition is that $r$ must be a multiple of 4. The possible values for $r$ are $0, 1, 2, \dots, 124$. We need to find the number of multiples of 4 in this range. Let $r = 4k$. $0 \le 4k \le 124$ $0 \le k \le \frac{124}{4}$ $0 \le k \le 31$ The possible integer values for $k$ are $0, 1, 2, \dots, 31$. The number of these values is $31 - 0 + 1 = 32$. Therefore, there are 32 rational terms.