मराठी

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

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प्रश्न

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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उत्तर

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

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Introduction of Binomial Theorem
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 18: Binomial Theorem - Exercise 18.2 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 18 Binomial Theorem
Exercise 18.2 | Q 10 | पृष्ठ ३८

संबंधित प्रश्‍न

Using binomial theorem, write down the expansions  :

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\[= ^{5}{}{C}_0 (2x )^5 (3y )^0 +^{5}{}{C}_1 (2x )^4 (3y )^1 + ^{5}{}{C}_2 (2x )^3 (3y )^2 + ^{5}{}{C}_3 (2x )^2 (3y )^3 + ^{5}{}{C}_4 (2x )^1 (3y )^4 +^{5}{}{C}_5 (2x )^0 (3y )^5\]

\[= 32 x^5 + 5 \times 16 x^4 \times 3y + 10 \times 8 x^3 \times 9 y^2 + 10 \times 4 x^2 \times 27 y^3 + 5 \times 2x \times 81 y^4 + 243 y^5 \]
\[ = 32 x^5 + 240 x^4 y + 720 x^3 y^2 + 1080 x^2 y^3 + 810x y^4 + 243 y^5 \]

 

 


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