मराठी

The Term Without X in the Expansion of ( 2 X − 1 2 X 2 ) 12 is (A) 495 (B) −495 (C) −7920 (D) 7920 - Mathematics

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प्रश्न

The term without x in the expansion of \[\left( 2x - \frac{1}{2 x^2} \right)^{12}\] is 

 

पर्याय

  • 495

  • −495

  • −7920

  •  7920

     
MCQ

उत्तर

 7920

\[\text{ Suppose the } (r + 1)\text{ th term in the given expansion is independent of  } x . \]

\[\text{ Then, we have: }  \]

\[ T_{r + 1} = ^{12}{}{C}_r (2x )^{12 - r} \left( \frac{- 1}{2 x^2} \right)^r \]

`= ( - 1 )^r "^12 C _r 2^{12 - 2r} x^{12 - r - 2r}`

\[\text{ For this term to be independent of x, we must have: } \]

\[12 - 3r = 0\]

\[ \Rightarrow r = 4\]

\[ \therefore \text{ Required term: } \]

`( - 1 )^4 "^12C_4 2^{12 - 8}`

\[ = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} \times 16\]

\[ = 7920\]

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Introduction of Binomial Theorem
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पाठ 18: Binomial Theorem - Exercise 18.4 [पृष्ठ ४६]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 18 Binomial Theorem
Exercise 18.4 | Q 2 | पृष्ठ ४६

संबंधित प्रश्‍न

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

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