Question
${(3 + 2x)^{50}}$ के विस्तार में महत्तम पद है, जहाँ $x = \frac{1}{5}$
$\therefore \,\,\,\frac{{{T_{r + 1}}}}{{{T_r}}} \ge 1\,\,\, \Rightarrow 102 - 2r \ge 15r \Rightarrow r \le 6$
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