<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] Reference that $$(a + b)^n = {n \choose 0}a^nb^0 + {n \choose 1}a^{n-1}b^1 + \cdots + {n \choose n-1}ab^{n-1} + {n \choose n}a^0b^n.$$
So we want $a = 2$ and $b = {x \over 3}$. So we are considering the terms $\displaystyle {n \choose 7}a^{n - 7}b^7$ and $\displaystyle {n\choose 8}a^{n-8}b^8.$ So, $${n \choose 7}a^{n - 7}b^7 = {n\choose 8}a^{n-8}b^8$$ $${n! \over 7!(n - 7)!}a^{n-7}b^7 = {n! \over 8!(n-8)!}a^{n-8}b^8$$ $${a^{n-8}a \over 7!(n-7)(n-8)!} = {a^{n-8}b \over 8\times 7!(n-8)!}$$ $${a \over n-7} = {b \over 8}$$ $$n-7 = {8a \over b}$$ $$n = {48 \over x} + 7$$ Can you see where it comes in now?
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