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7.Binomial Theorem
easy
${\left( {{x^2} - \frac{{3\sqrt 3 }}{{{x^3}}}} \right)^{10}}$ ના વિસ્તરણમાં અચળપદ મેળવો.
A
$153090$
B
$150000$
C
$150090$
D
$153180$
Solution
(a) ${T_{r + 1}} = {}^{10}{C_r}{({x^2})^{10 – r}}{\left( {\frac{{ – 3\sqrt 3 }}{{{x^3}}}} \right)^r}$
For term independent of $x$, $20 – 2r – 3r = 0 \Rightarrow r = 4$
$ \therefore {T_{4 + 1}} = {}^{10}{C_4}{( – 3)^4}{(\sqrt 3 )^4} = 153090.$
Standard 11
Mathematics
