7. binomial theoram — ગણિત ધોરણ 11 સાયન્સ — Question

${T_{r + 1}} = {\,^9}{C_r}{({x^2})^{9 - r}}{\left( { - \frac{1}{{3x}}} \right)^r}$

$ = {\,^9}{C_r}{x^{18 - 2r}}\frac{{{{( - 1)}^r}}}{{{3^r}}}{x^{ - r}}$

It is independent of $x$.

$\therefore  18 - 3r = 0 \Rightarrow r = 6$

 $\therefore {T_7} = {\,^9}{C_6}{x^{18 - 12}}\frac{{{{( - 1)}^6}}}{{{3^6}}}{x^{ - 6}} = {\,^9}{C_6}\frac{{{{( - 1)}^6}}}{{36}} = \frac{{28}}{{243}}$