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

Answer 1

(d) 101

$$\begin{gathered} \mathrm{The}\mathrm{general}\mathrm{term}{T}_{r+1}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}\mathrm{is}\mathrm{given}\mathrm{by} \\ {}^{600}C_r({17}^{1/3}{)}^{600-r}({35}^{1/2}x{)}^r \\ ={}^{600}C_r{17}^{200-r/3}\times {35}^{r/2}{x}^r \\ \mathrm{Now},T_{r+1}\mathrm{is}\mathrm{an}\mathrm{integer}\mathrm{if}\frac{\mathit{}r}{2}\mathrm{and}\frac{r}{3}\mathrm{are}\mathrm{integers}\mathrm{for}\mathrm{all}0\le r\le 600 \\ \mathrm{Thus},\mathrm{we}\mathrm{have} \\ r=0,6,12,...600(\mathrm{Multiples}\mathrm{of}6) \\ \mathrm{Since},\mathrm{It}\mathrm{is}\mathrm{an}\mathrm{A}.\mathrm{P} \\ \mathrm{So},600=0+\left(n-1\right)6 \\ \Rightarrow n=101 \\ \mathrm{Hence},\mathrm{there}\mathrm{are}101\mathrm{terms}\mathrm{with}\mathrm{integral}\mathrm{coefficients}. \end{gathered}$$