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Evaluate the (Vi) ( 2 + √ 3 ) 7 + ( 2 − √ 3 ) 7 - Mathematics

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प्रश्न

Evaluate the

(vi)  \[\left( 2 + \sqrt{3} \right)^7 + \left( 2 - \sqrt{3} \right)^7\]

उत्तर

(vi)  \[(2 + \sqrt{3} )^7 + (2 - \sqrt{3} )^7 \]
\[ = 2[ ^{7}{}{C}_0 \times 2^7 \times (\sqrt{3} )^0 + ^{7}{}{C}_2 \times 2^5 \times (\sqrt{3} )^2 + ^{7}{}{C}_4 \times 2^3 \times (\sqrt{3} )^4 +^{7}{}{C}_6 \times 2^1 \times (\sqrt{3} )^6 ]\]
\[ = 2[128 + 21 \times 32 \times 3 + 35 \times 8 \times 9 + 7 \times 2 \times 27]\]
\[ = 2[128 + 2016 + 2520 + 378]\]
\[ = 2 \times 5042 = 10084\] 

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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]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.1 | Q 2.06 | पृष्ठ ११

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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