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The Number of Terms with Integral Coefficients in the Expansion of ( 17 1 / 3 + 35 1 / 2 X ) 600 is (A) 100 (B) 50 (C) 150 (D) 101

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प्रश्न

The number of terms with integral coefficients in the expansion of \[\left( {17}^{1/3} + {35}^{1/2} x \right)^{600}\] is

 

विकल्प

  • 100

  •  50

  •  150

  • 101

     
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उत्तर

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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अध्याय 18: Binomial Theorem - Exercise 18.4 [पृष्ठ ४८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.4 | Q 31 | पृष्ठ ४८

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