Q115 / 225 · JEE Main 2003 · Mathematics

The number of integral terms in the expansion of $(\sqrt{3}+8 \sqrt{5})^{256}$ is

  • A
    35
  • B
    32
  • Correct answer
    C
    33
  • D
    34

Solution

$\mathrm{T}_{\mathrm{r}+1}={ }^{256} \mathrm{C}_{\mathrm{r}}(\sqrt{3})^{256-\mathrm{r}}(\sqrt[8]{5})^{\mathrm{r}}={ }^{256} \mathrm{C}_{\mathrm{r}}(3)^{\frac{256-\mathrm{r}}{2}}(5)^{\mathrm{r} / 8}$ Terms will be integral if $\frac{256-\mathrm{r}}{2} \& \frac{\mathrm{r}}{8}$ both are +ve integer. As $0 \leq \mathrm{r} \leq 256 \therefore \mathrm{r}=0,8,16,24, \ldots \ldots \ldots \ldots 256$ For above values of $\mathrm{r},\left(\frac{256-\mathrm{r}}{2}\right)$ is also an integer.

Hence, the correct answer is option: 'C'.

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