The greatest term in the expansion of sqrt{3} | vedclass

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Question

(A) Let $T_{r+1}$ be the greatest term in the expansion of $\sqrt{3} \left( 1 + \frac{1}{\sqrt{3}} \right)^{20}$.We have $T_{r+1} = \sqrt{3} \cdot {}^

Vedclass · Accepted Answer

$\frac{25840}{9}$

Vedclass · Answer

(A) Let $T_{r+1}$ be the greatest term in the expansion of $\sqrt{3} \left( 1 + \frac{1}{\sqrt{3}} ight)^{20}$.We have $T_{r+1} = \sqrt{3} \cdot {}^{20}C_r \left( \frac{1}{\sqrt{3}} ight)^r$ and $T_r = \sqrt{3} \cdot {}^{20}C_{r-1} \left( \frac{1}{\sqrt{3}} ight)^{r-1}$.For the greatest term,we consider the ratio $\frac{T_{r+1}}{T_r} \ge 1$.$\frac{T_{r+1}}{T_r} = \frac{{}^{20}C_r}{{}^{20}C_{r-1}} \cdot \frac{1}{\sqrt{3}} = \frac{20-r+1}{r} \cdot \frac{1}{\sqrt{3}} \ge 1$.$21 - r \ge r\sqrt{3} \Rightarrow 21 \ge r(\sqrt{3} + 1)$.$r \le \frac{21}{\sqrt{3} + 1} = \frac{21(\sqrt{3}-1)}{3-1} = \frac{21(1.732 - 1)}{2} = \frac{21 	imes 0.732}{2} \approx 7.686$.Since $r$ must be an integer,the greatest term occurs at $r = 7$,which is $T_{7+1} = T_8$.$T_8 = \sqrt{3} \cdot {}^{20}C_7 \left( \frac{1}{\sqrt{3}} ight)^7 = \frac{{}^{20}C_7}{(\sqrt{3})^6} = \frac{77520}{27} = \frac{25840}{9}$.

The greatest term in the expansion of $\sqrt{3} \left( 1 + \frac{1}{\sqrt{3}} \right)^{20}$ is

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