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Q. If the coefficients of $x^7$ and $x^8$ in $\left(2+\frac{x}{3}\right)^{n}$ are equal, then $n$ is

Binomial Theorem

Solution:

$T_{8} = \,{}^{n}C_{7}\left(2\right)^{n-7}\left(\frac{x}{3}\right)^{7}$
$= \,{}^{n}C_{7} \frac{2^{n-7}}{3^{7}}x^{7}$
and $T_{9} = \,{}^{n}C_{8}\left(2\right)^{n-8}\left(\frac{x}{3}\right)^{8}$
$= \,{}^{n}C_{8} \frac{2^{n-8}}{3^{8}}x^{8}$
Now, it is given that $\,{}^{n}C_{7} \frac{\left(2\right)^{n-7}}{3^{7}} = \,{}^{n}C_{8} \frac{2^{n-8}}{3^{8}}$
$\Rightarrow \frac{8}{n-7}=\frac{1}{6}$
$\Rightarrow n=55$