# The number of terms with integral coefficients in the expansion of

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.