Question:
The number of terms with integral coefficients in the expansion of $\left(17^{1 / 3}+35^{1 / 2} x\right)^{600}$ is
(a) 100
(b) 50
(c) 150
(d) 101
Solution:
(d) 101
The general term $T_{r+1}$ in the given expansion is given by
${ }^{600} C_{r}\left(17^{1 / 3}\right)^{600-r}\left(35^{1 / 2} x\right)^{r}$
$={ }^{600} C_{r} 17^{200-r / 3} \times 35^{r / 2} x^{r}$
Now, $T_{r+1}$ is an integer if $\frac{r}{2}$ and $\frac{r}{3}$ are integers for all $0 \leq r \leq 600$
Thus, we have
$r=0,6,12, \ldots 600$ (Multiples of 6 )
Since, It is an A.P
So, $600=0+(n-1) 6$
$\Rightarrow n=101$
Hence, there are 101 terms with integral coefficients.