Question:
If $\mathrm{n}$ is the number of irrational terms in the expansion of $\left(3^{1 / 4}+5^{1 / 8}\right)^{60}$, then $(n-1)$ is divisible by :
Correct Option: 1
Solution:
$\left(3^{1 / 4}+5^{1 / 8}\right)^{60}$
${ }^{60} C_{r}\left(3^{1 / 4}\right)^{60-r} \cdot\left(5^{1 / 8}\right)^{r}$
${ }^{60} C_{r}(3)^{\frac{60-r}{4}} .5^{\frac{r}{8}}$
For rational terms.
$\frac{\mathrm{r}}{8}=\mathrm{k} ; \quad 0 \leq \mathrm{r} \leq 60$
$0 \leq 8 \mathrm{k} \leq 60$
$0 \leq \mathrm{k} \leq \frac{60}{8}$
$0 \leq \mathrm{k} \leq 7.5$
$\mathrm{k}=0,1,2,3,4,5,6,7$
$\frac{60-8 \mathrm{k}}{4}$ is always divisible by 4 for all value
of $\mathrm{k}$
Total rational terms $=8$
Total terms $=61$
irrational terms $=53$
$n-1=53-1=52$
52 is divisible by 26 .