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Evaluate the (I) ( √ X + 1 + √ X − 1 ) 6 + ( √ X + 1 − √ X − 1 ) 6 - Mathematics

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प्रश्न

Evaluate the 

(i)\[\left( \sqrt{x + 1} + \sqrt{x - 1} \right)^6 + \left( \sqrt{x + 1} - \sqrt{x - 1} \right)^6\]

 

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उत्तर

(i) \[(\sqrt{x + 1} + \sqrt{x - 1} )^6 + (\sqrt{x + 1} - \sqrt{x - 1} )^6 \]
\[ = 2[ ^{6}{}{C}_0 (\sqrt{x + 1} )^6 (\sqrt{x - 1} )^0 + ^{6}{}{C}_2 (\sqrt{x + 1} )^4 (\sqrt{x - 1} )^2 +^{6}{}{C}_4 (\sqrt{x + 1} )^2 (\sqrt{x - 1} )^4 + ^{6}{}{C}_6 (\sqrt{x + 1} )^0 (\sqrt{x - 1} )^6 ]\]
\[ = 2[(x + 1 )^3 + 15(x + 1 )^2 (x - 1) + 15(x + 1)(x - 1 )^2 + (x - 1 )^3 \]
\[ = 2[ x^3 + 1 + 3x + 3 x^2 + 15( x^2 + 2x + 1)(x - 1) + 15(x + 1)( x^2 + 1 - 2x) + x^3 - 1 + 3x - 3 x^2 ]\]
\[ = 2[2 x^3 + 6x + 15 x^3 - 15 x^2 + 30 x^2 - 30x + 15x - 15 + 15 x^3 + 15 x^2 - 30 x^2 - 30x + 15x + 15]\]
\[ = 2[32 x^3 - 24x]\]
\[ = 16x[4 x^2 - 3]\]

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Introduction of Binomial Theorem
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Binomial Theorem - Exercise 18.1 [पृष्ठ ११]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.1 | Q 2.01 | पृष्ठ ११

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