If n is the number of irrational terms in the expansion of

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 :

  1. 26

  2. 30

  3. 8

  4. 7


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 .

 

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