7.Binomial Theorem
medium

यदि  ${(1 + x)^m}$ के द्विपद प्रसार में तृतीय पद  $ - \frac{1}{8}{x^2}$ है, तब $m$ का परिमेय मान है

A

$2$

B

$1/2$

C

$3$

D

$4$

Solution

${(1 + x)^m} = 1 + mx + \frac{{m(m – 1)}}{{2!}}{x^2} + …$

परिकल्पना से $\frac{{m(m – 1)}}{2}{x^2} =  – \frac{1}{8}{x^2}$

 $⇒4{m^2} – 4m =  – 1$

 $⇒ {(2m – 1)^2} = 0 $

$\Rightarrow m = \frac{1}{2}$.

Standard 11
Mathematics

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