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JEE Mains · Maths · STD 11 - 7. binomial theoram

\(\left( {{2^{1/3}} + \frac{1}{{2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}\) ના વિસ્તરણમાં પહેલેથી \(5^{th}\) માં પદ અને છેલ્લેથી \(5^{th}\) માં પદનો ગુણોત્તર મેળવો.

  1. A \(1:2{\left( 6 \right)^{\frac{1}{3}}}\)
  2. B \(1:4{\left( 16 \right)^{\frac{1}{3}}}\)
  3. C \(4{\left( {36} \right)^{\frac{1}{3}}}\,:\,1\)
  4. D \(2{\left( {36} \right)^{\frac{1}{3}}}\,:\,1\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4{\left( {36} \right)^{\frac{1}{3}}}\,:\,1\)

Step-by-step Solution

Detailed explanation

\frac{{{5^{{\text{th}}}}{\text{ term from begining }}}}{{{5^{{\text{th}}}}{\text{ term from end }}}} = \frac{{10{{\text{C}}_4}{{\left( {\frac{1}{{2\left( {{3^{1/3}}} \right)}}} \right)}^4}{2^{6/3}}}}{{10{{\text{C}}_4}{{(2)}^{4/3}}{{\left( {\frac{1}{{2\left( {{3^{1/3}}}…

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