If $\left(2+\frac{x}{3}\right)^{55}$ is expanded in the ascending powers of $x$ and the coefficients of…
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:
$7^{\text {th }}$ and $8^{\text {th }}$
$8^{\text {th }}$ and $9^{\text {th }}$
$28^{\text {th }}$ and $29^{\text {th }}$
$27^{\text {th }}$ and $28^{\text {th }}$
Solution
Let $r^{\text {th }}$ and $(r+1)^{\text {th }}$ term has equal coefficient
$
\begin{aligned}
&\left(2+\frac{x}{3}\right)^{55}=2^{55}\left(1+\frac{x}{6}\right)^{55} \\
&r^{\text {th }} \text { term }=2^{5555} C_r\left(\frac{x}{6}\right)^r
\end{aligned}
$
Coefficient of $x^r$ is $2^{55}{ }^{55} C_r \frac{1}{6^r}$
$
(r+1)^{\text {th }} \text { term }=2^{5555} C_{r+1}\left(\frac{x}{6}\right)^{r+1}
$
Coefficient of $x^{r+1}$ is $2^{55}{ }^{55} C_{r+1} \cdot \frac{1}{6^{r+1}}$
Both coefficients are equal
$
2^{5555} C_r \frac{1}{6^r}=2^{5555} C_{r+1} \frac{1}{6^{r+1}}
$
$
\begin{aligned}
&\frac{1}{|r| 55-r}=\frac{1}{\lfloor r+1 \mid 54-r} \cdot \frac{1}{6} \\
&6(r+1)=55-r \\
&6 r+6=55-r \\
&7 r=49 \\
&r=7 \\
&(r+1)=8
\end{aligned}
$
Coefficient of $7^{\text {th }}$ and $8^{\text {th }}$ terms are equal