Question

In the expansion of ${(a^{1/3}+b^{1/9})}^{6561},$ where $a,b$ are distinct prime numbers, if the number of irrational terms is $N,$ then the value of $\frac{N-32}{100}$ is

Answer 1

$T_{r+1}={}^{6561}{C}_r\left(a^{1/3}{)}^{6561-r}\right(b{)}^{r/9}$
$={}^{6561}{C}_r\left(a{)}^{(6561-r)/3}\right(b{)}^{r/9}$
$T_{r+1}$ will be integer if $\frac{r}{9}$ and $\frac{6561-r}{3}$
both are integers.
$r=0,9,18,\dots ,6561.$
Number of values of $r=\frac{6561}{9}+1=730.$
So, number of irrational terms $=6562-730=5832$
Now, $\frac{5832-32}{100}=58$