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If the coefficient of $x^{5}$ and $x^{6}$ in $\left(2+\frac{x}{3}\right)^{n}$ are equal, then $n$ is
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Verified Answer
The correct answer is:
41
Given, expansion is $\left(2+\frac{x}{3}\right)^{n}$
Let $t_{r+1}$ be general term.
Then, $t_{r+1}={ }^{n} C_{r} 2^{n-r}\left(\frac{x}{3}\right)^{r}={ }^{n} C_{r} 2^{n-r} \cdot 3^{-r} x^{r}$
Since, coefficient of $x^{5}$ and $x^{6}$ are equal.
$$
\begin{aligned}
&\therefore \quad{ }^{n} C_{6} 2^{n-6} 3^{-6}={ }^{n} C_{5} 2^{n-5} 3^{-5} \\
&\Rightarrow \quad \frac{{ }^{n} C_{6}}{{ }^{n} C_{5}}=2 \times 3 \Rightarrow \frac{n ! \times 5 ! \times(n-5) !}{(n-6) ! \times 6 ! \times n !}=6 \\
&\Rightarrow \quad \frac{n-5}{6}=6 \Rightarrow n-5=36 \Rightarrow n=41
\end{aligned}
$$
Let $t_{r+1}$ be general term.
Then, $t_{r+1}={ }^{n} C_{r} 2^{n-r}\left(\frac{x}{3}\right)^{r}={ }^{n} C_{r} 2^{n-r} \cdot 3^{-r} x^{r}$
Since, coefficient of $x^{5}$ and $x^{6}$ are equal.
$$
\begin{aligned}
&\therefore \quad{ }^{n} C_{6} 2^{n-6} 3^{-6}={ }^{n} C_{5} 2^{n-5} 3^{-5} \\
&\Rightarrow \quad \frac{{ }^{n} C_{6}}{{ }^{n} C_{5}}=2 \times 3 \Rightarrow \frac{n ! \times 5 ! \times(n-5) !}{(n-6) ! \times 6 ! \times n !}=6 \\
&\Rightarrow \quad \frac{n-5}{6}=6 \Rightarrow n-5=36 \Rightarrow n=41
\end{aligned}
$$
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