If {left( {2 + frac{x}{3}} right)^{55}} is ex | vedclass

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(A) Let the $(r+1)^{	ext{th}}$ and $(r+2)^{	ext{th}}$ terms have equal coefficients.${\left(2+\frac{x}{3}ight)^{55} = 2^{55}\left(1+\frac{x}{6}i

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$8^{th}$ and $9^{th}$

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(A) Let the $(r+1)^{	ext{th}}$ and $(r+2)^{	ext{th}}$ terms have equal coefficients.${\left(2+\frac{x}{3}ight)^{55} = 2^{55}\left(1+\frac{x}{6}ight)^{55}}$The $(r+1)^{	ext{th}}$ term is $2^{55} \cdot {}^{55}C_r \left(\frac{x}{6}ight)^r$. The coefficient of $x^r$ is $2^{55} \cdot {}^{55}C_r \cdot \frac{1}{6^r}$.The $(r+2)^{	ext{th}}$ term is $2^{55} \cdot {}^{55}C_{r+1} \left(\frac{x}{6}ight)^{r+1}$. The coefficient of $x^{r+1}$ is $2^{55} \cdot {}^{55}C_{r+1} \cdot \frac{1}{6^{r+1}}$.Equating the coefficients:${}^{55}C_r \cdot \frac{1}{6^r} = {}^{55}C_{r+1} \cdot \frac{1}{6^{r+1}}$${}^{55}C_r = {}^{55}C_{r+1} \cdot \frac{1}{6}$$6 \cdot {}^{55}C_r = {}^{55}C_{r+1}$$6 \cdot \frac{55!}{r!(55-r)!} = \frac{55!}{(r+1)!(54-r)!}$$6 \cdot \frac{1}{r!(55-r)(54-r)!} = \frac{1}{(r+1)r!(54-r)!}$$\frac{6}{55-r} = \frac{1}{r+1}$$6(r+1) = 55-r$$6r + 6 = 55 - r$$7r = 49$$r = 7$Thus,the terms are $(r+1)^{	ext{th}} = 8^{	ext{th}}$ and $(r+2)^{	ext{th}} = 9^{	ext{th}}$.

If ${\left( {2 + \frac{x}{3}} \right)^{55}}$ is expanded in the ascending powers of $x$ and the coefficients of powers of $x$ in two consecutive terms of the expansion are equal,then these terms are

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