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7.Binomial Theorem
easy
${\left( {2x - \frac{1}{{2{x^2}}}} \right)^{12}}$ के प्रसार में $x$ से स्वतंत्र पद है
A
$-7930$
B
$-495$
C
$495$
D
$7920$
Solution
${(x)^{12 – r}}{\left( {\frac{1}{{{x^2}}}} \right)^r} = {x^0}\,\, \Rightarrow {x^{12 – 3r}} = {x^0} \Rightarrow r = 4$
अत: अभीष्ट पद $^{12}{C_4}{2^8}{\left( { – \frac{1}{2}} \right)^4} = 7920$.
Standard 11
Mathematics