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7.Binomial Theorem
easy
${\left( {x - \frac{1}{{2x}}} \right)^8}$ के विस्तार में ${x^2}$ का गुणांक होगा
A
$\frac{1}{7}$
B
$\frac{{ - 1}}{7}$
C
$-7$
D
$7$
Solution
${T_{r + 1}}{ = ^8}{C_r}{(x)^{8 – r}}{\left( { – \frac{1}{{2x}}} \right)^r}$
${ = ^8}{C_r}{\left( {\frac{{ – 1}}{2}} \right)^r}{x^{8 – r – r}}$
${x^2}$ के गुणांक के लिए $8 – 2r = 2$ ==> $r = 3$
${x^2}$ का गुणांक ${ = ^8}{C_3}{\left( {\frac{{ – 1}}{2}} \right)^3}$= $ – \frac{{8!}}{{5!\,3!}}\frac{1}{{{2^3}}} = – 7$.
Standard 11
Mathematics
