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The greatest term in the expansion of $\sqrt{3}\left(1+\frac{1}{\sqrt{3}}\right)^{20}$ is
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The correct answer is:
$\frac{25840}{9}$
Let $(r+1)^{\text {th }}$ term be the greatest term. Then
$\begin{aligned} & T_{r+1}=\sqrt{3} \cdot{ }^{20} C_r\left(\frac{1}{\sqrt{3}}\right)^r \text { and } T_r=\sqrt{3} \cdot{ }^{20} C_{r-1}\left(\frac{1}{\sqrt{3}}\right)^{r-1} \\ & \text { Now } \frac{T_{r+1}}{T_r}=\frac{20-r+1}{r}\left(\frac{1}{\sqrt{3}}\right)\end{aligned}$
$\begin{aligned} & \therefore T_{r+1} \geq T_r \Rightarrow 20-r+1 \geq \sqrt{3} r \\ & \Rightarrow 21 \geq r(\sqrt{3}+1) \Rightarrow r \leq \frac{21}{\sqrt{3}+1} \Rightarrow r \leq 7.686 \Rightarrow r=7\end{aligned}$
Hence the greatest term is $T_8=\sqrt{3}{ }^{20} C_7\left(\frac{1}{\sqrt{3}}\right)^7=\frac{25840}{9}$
$\begin{aligned} & T_{r+1}=\sqrt{3} \cdot{ }^{20} C_r\left(\frac{1}{\sqrt{3}}\right)^r \text { and } T_r=\sqrt{3} \cdot{ }^{20} C_{r-1}\left(\frac{1}{\sqrt{3}}\right)^{r-1} \\ & \text { Now } \frac{T_{r+1}}{T_r}=\frac{20-r+1}{r}\left(\frac{1}{\sqrt{3}}\right)\end{aligned}$
$\begin{aligned} & \therefore T_{r+1} \geq T_r \Rightarrow 20-r+1 \geq \sqrt{3} r \\ & \Rightarrow 21 \geq r(\sqrt{3}+1) \Rightarrow r \leq \frac{21}{\sqrt{3}+1} \Rightarrow r \leq 7.686 \Rightarrow r=7\end{aligned}$
Hence the greatest term is $T_8=\sqrt{3}{ }^{20} C_7\left(\frac{1}{\sqrt{3}}\right)^7=\frac{25840}{9}$
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