7.Binomial Theorem
easy

${\left( {ax - \frac{1}{{b{x^2}}}} \right)^{11}}$ के प्रसार में ${x^{ - 7}}$ का गुणांक होगा

A

$\frac{{462{a^6}}}{{{b^5}}}$

B

$\frac{{462{a^5}}}{{{b^6}}}$

C

$\frac{{ - 462{a^5}}}{{{b^6}}}$

D

$\frac{{ - 462{a^6}}}{{{b^5}}}$

(IIT-1967)

Solution

पदों की संख्या के लिए

$(11 – r)(1) + r( – 2) =  – 7 $

$\Rightarrow 11 – r – 2r =  – 7$

$ \Rightarrow r = 6$

अत: $x^{-7}$  का गुणांक  $^{11}{C_6}{(a)^5}{\left( { – \frac{1}{b}} \right)^6} = \frac{{462}}{{{b^6}}}{a^5}$

 

Standard 11
Mathematics

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