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