The number of terms with integral coefficients in the expansion of $\Big(17^{\frac{1}{3}}+35^{\frac{1}{2}}\text{x}\Big)^{600}$ is:

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Aicharya · Accepted Answer

Solution: The general term T r+1 in the given expansion is given by ${^\text{600}}\text{C}_{\text{r}}(17^{\frac{1}{3}})^{600-\text{r}}\Big(35^{\frac{1}{2}}\text{x}\Big)^{\text{r}}$ $={^\text{600}}\text{C}_{\text{r}}(17^{\frac{1}{3}})^{200-\frac{\text{r}}{3}}\times\Big(35^{\frac{\text{r}}{2}}\text{x}\Big)^{\text{r}}$ Now, T r+1 is an integer if $\frac{\text{r}}{2}$ and $\frac{\text{r}}{2}$ are integers for all $0\leq\text{r}\leq600$ Thus, we have r = 0, 6, 12, ....600 Since, It is an A.P So, $600 = 0 + (\text{n} - 1)6$ $\Rightarrow \text{n}=101$ Hence, there are 101 terms with integral coefficients.

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

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