7.Binomial Theorem
hard

${\left( {1 - 2\sqrt x } \right)^{50}}$ના દ્ઘિપદી વિસ્તરણમાં $x $ ની પૂર્ણાક ઘાતાંકના સહગુણકોનો સરવાળો . . . . . . . . . . થાય. 

A

$\frac{1}{2}\left( {{2^{50}} + 1} \right)$

B

$\;\frac{1}{2}\left( {{3^{50}} + 1} \right)$

C

$\;\frac{1}{2}\left( {{3^{50}}} \right)$

D

$\;\frac{1}{2}\left( {{3^{50}} - 1} \right)$

(JEE MAIN-2015)

Solution

${\left( {1 – 2\sqrt x } \right)^{50}} = {^{50}}{c_0} – {^{50}}{c_1}{\left( {2\sqrt x } \right)^1}$${\left( {1 – 2\sqrt x } \right)^{50}} = {^{50}}{c_0} – {^{50}}{c_1}{\left( {2\sqrt x } \right)^1}$$ + {^{50}}{c_4}\left( {2\sqrt x } \right)4…..$

$s = {^{50}}{c_0} + {^{50}}{c_2} \cdot {2^2} + {^{50}}{c_4}{\left( 2 \right)^4} + …. + {^{50}}c_{502}^{50}$

${\left( {1 + x} \right)^{50}} = 1 + {^{50}}{C_1}{x^1} + {^{50}}{c_2}{x^2} + …$

$x = 2, – 2$

${\left( {1 + 2} \right)^{50}} = 1 + {^{50}}{c_1}\left( 2 \right) + {^{50}}{c_2}{\left( 2 \right)^2} + ..$

${\left( {1 – 2} \right)^{50}} = 1 – {^{50}}{c_1}\left( 2 \right) + $${^{50}}{c_2}{\left( 2 \right)^2} – {^{50}}{C_3}{\left( 2 \right)^3} + …..$

${3^{50}} + 1$$ = 2\left[ {{^{50}}{C_0} + {^{50}}{c_2}{{\left( 2 \right)}^2} + {^{50}}{c_4}{{\left( 2 \right)}^4} + ..} \right]$

$\frac{{{3^{50}} + 1}}{2}$$ = {^{50}}{C_0} + {^{50}}{c_2}\left( 2 \right) + {^{50}}{c_4}{\left( 2 \right)^4} + …$

Standard 11
Mathematics

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