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7.Binomial Theorem
easy
${\left( {\frac{{{x^2}}}{2} - \frac{2}{x}} \right)^9}$ के विस्तार में ${x^{ - 9}}$ का गुणांक होगा
A
$512$
B
$-512$
C
$521$
D
$251$
Solution
यहाँ ${T_{r + 1}} = {}^{\rm{9}}{C_r}{\left( {\frac{{{x^2}}}{2}} \right)^{9 – r}}{\left( {\frac{{ – 2}}{x}} \right)^r}$
= ${}^9{C_r}\frac{{{x^{18 – 3r}}{{( – 2)}^r}}}{{{2^{9 – r}}}},$ ${x^{ – 9}}$ के गुणांक के लिए $18 -3r = -9$ अर्थात् $r = 9$ ${x^{ – 9}}$ का गुणांक
= ${}^9{C_9}\frac{{{{( – 2)}^9}}}{{{2^0}}} = – {2^9} = – 512$.
Standard 11
Mathematics
