${\left( {\frac{{3{x^2}}}{2} + \frac{1}{{3x}}} \right)^9}$ के प्रसार में व्यापक पद

${T_{r + 1}} = {\,^9}{C_r}.{\left( {\frac{{3{x^2}}}{2}} \right)^{9 – r}}{\left( { – \frac{1}{{3x}}} \right)^r}$$ = {\,^9}{C_r}{\left( {\frac{3}{2}} \right)^{9 – r}}{\left( { – \frac{1}{3}} \right)^r}{x^{18 – 3r}}$

$x$ से स्वतंत्र पद के लिए $18 -3r = 0$ ==> $r = 6$

अत: ${T_{6 + 1}} = {\,^9}{C_6}{\left( {\frac{3}{2}} \right)^{9 – 6}}{\left( { – \frac{1}{3}} \right)^6} = {\,^9}{C_3}.\frac{1}{{{6^3}}}$