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7.Binomial Theorem
easy
The term independent of $x$ in the expansion ${\left( {{x^2} - \frac{1}{{3x}}} \right)^9}$ is
A
$\frac{{28}}{{81}}$
B
$\frac{{28}}{{243}}$
C
$ - \frac{{28}}{{243}}$
D
$ - \frac{{28}}{{81}}$
Solution
(b) In ${\left( {{x^2} – \frac{1}{{3x}}} \right)^9},$
${T_{r + 1}} = {\,^9}{C_r}{({x^2})^{9 – r}}{\left( { – \frac{1}{{3x}}} \right)^r}$
$ = {\,^9}{C_r}{x^{18 – 2r}}\frac{{{{( – 1)}^r}}}{{{3^r}}}{x^{ – r}}$
It is independent of $x$.
$\therefore 18 – 3r = 0 \Rightarrow r = 6$
$\therefore {T_7} = {\,^9}{C_6}{x^{18 – 12}}\frac{{{{( – 1)}^6}}}{{{3^6}}}{x^{ – 6}} = {\,^9}{C_6}\frac{{{{( – 1)}^6}}}{{36}} = \frac{{28}}{{243}}$
Standard 11
Mathematics
