The largest term in the expansion of (3 + 2x) | vedclass

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(C) Given expression is $(3 + 2x)^{50} = 3^{50} (1 + \frac{2x}{3})^{50}$.Substituting $x = \frac{1}{5}$,we get $3^{50} (1 + \frac{2}{15})^{50} = 3^{50

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$7^{th}$

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(C) Given expression is $(3 + 2x)^{50} = 3^{50} (1 + \frac{2x}{3})^{50}$.Substituting $x = \frac{1}{5}$,we get $3^{50} (1 + \frac{2}{15})^{50} = 3^{50} (1 + \frac{2}{15})^{50}$.Let $T_{r+1}$ be the $(r+1)^{th}$ term.We know that $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot |\frac{a_2}{a_1}|$.Here $n=50$,$a_1=1$,$a_2=\frac{2}{15}$.So,$\frac{T_{r+1}}{T_r} = \frac{50-r+1}{r} \cdot \frac{2}{15} = \frac{51-r}{r} \cdot \frac{2}{15}$.For the term to be the largest,we set $\frac{T_{r+1}}{T_r} \ge 1$.$\frac{2(51-r)}{15r} \ge 1$ $\Rightarrow 102 - 2r \ge 15r$ $\Rightarrow 102 \ge 17r$ $\Rightarrow r \le 6$.Since $r=6$ satisfies the condition,$T_{6+1} = T_7$ is the largest term.

The largest term in the expansion of $(3 + 2x)^{50}$ where $x = \frac{1}{5}$ is

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