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Which Term in the Expansion of ⎧ ⎨ ⎩ ( X √ Y ) 1 / 3 + ( Y X 1 / 3 ) 1 / 2 ⎫ ⎬ ⎭ 21 Contains X and Y to One and the Same Power? - Mathematics

Which term in the expansion of \[\left\{ \left( \frac{x}{\sqrt{y}} \right)^{1/3} + \left( \frac{y}{x^{1/3}} \right)^{1/2} \right\}^{21}\]  contains x and y to one and the same power?

 

 

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Solution

Suppose Tr+1th term in the given expression contains x and y to one and the same power.
Then,  \[T_{r + 1} \text{ th term is} \]
\[ ^{21}{}{C}_r \left[ \left( \frac{x}{\sqrt{y}} \right)^{1/3} \right]^{21 - r} \left[ \left( \frac{y}{x^{1/3}} \right)^{{}^{1/2}} \right]^r \]
\[ =^{21}{}{C}_r \left( \frac{x^{(21 - r)/3}}{x^{r/6}} \right)\left( \frac{y^{r/2}}{y^{(21 - r)/6}} \right)\]
\[ = ^{21}{}{C}_r \left( x \right)^{7 - r/2} \left( y \right)^{2r/3 - 7/2} \]
\[\text{ Now, if x and y have the same power, then } \]
\[7 - \frac{r}{2} = \frac{2r}{3} - \frac{7}{2}\]
\[ \Rightarrow \frac{2r}{3} + \frac{r}{2} = 7 + \frac{7}{2}\]
\[ \Rightarrow \frac{7r}{6} = \frac{21}{2}\]
\[ \Rightarrow r = 9\]
\[\text{ Hence, the required term is the 10th term } \]

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 10 | Page 38
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