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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

Let $\alpha > 0$ be the smallest number such that the expansion of $(x^{2/3} + 2x^{-3})^{30}$ has a term $\beta x^{-\alpha}$,where $\beta \in \mathbb{N}$. Then $\alpha$ is equal to $.............$.

The term independent of $x$ in the expansion of $\left( \frac{1}{60} - \frac{x^8}{81} \right) \left( 2x^2 - \frac{3}{x^2} \right)^6$ is equal to

$A$ person takes an examination consisting of four papers,each with a maximum of $m$ marks. The number of ways in which one can obtain a total of $2m$ marks is

If for some positive integer $n,$ the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+5}$ are in the ratio $5: 10: 14,$ then the largest coefficient in this expansion is

Find the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module