<s>[INST] Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal 

Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.



My attempt:
$\displaystyle \binom{n}{7}=\binom{n}{8} $
$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$
$$ \frac{6}{7!} = \frac{n-7}{40320} $$
$$ n-7 = 48 $$
$$ n=55 $$
 [/INST] $\displaystyle \left(2+\frac{x}{3}\right)^{n}=\sum_{i=0} ^n {n \choose k}({\frac{x}{3}})^i 2^{n-i}$
Now for $i =7\implies{n \choose 7}({\frac{x}{3}})^7 2^{n-7}$  
and for $i =8\implies{n \choose 8}({\frac{x}{3}})^8 2^{n-8}$  
now to get the coefficients of $x^8$ & $x^7$ equal.   $\implies$ 
${n \choose 7}({\frac{1}{3}})^7 2^{n-7}={n \choose 8}({\frac{1}{3}})^8 2^{n-8} \implies \frac{1}{n-7}=\frac{1}{8}* \frac{1}{3}*\frac{1}{2} \implies n=55$
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