binomial-coefficients question $(\frac{x\sqrt[3]{x}}{2}+\frac{1}{\sqrt[15]{x^28}})^n$

Question

I need to find the The coefficient of the free element (when the power is 0) for

$\left(\frac{x\sqrt[3]{x}}{2}+\frac{1}{\sqrt[15]{x^{28}}}\right)^n$

such that the sum of first coefficient and the coefficient second from the end is 13.

I used binom newton thoerm

where $\sum_{i}^{n}\binom{n}{i}(\frac{1}{2})^{i}*1^{n-i}*x^{\frac{3}{4}(i)-\frac{28}{15}(n-i)}$

I think the first coefficient is when $i=0$ and second from the end is $i=n-1$

I tried to substitute the $i's$ and equaling the sum of coefficients to 13 but I dont think that's the way..

Answer 1

The "sum of coefficients" likely refers to the sum of just the binomial coefficients $\binom ni$. Otherwise the sum of the first and penultimate coefficient is $\binom n0+\binom n{n-1}\cdot2^{1-n}=1+n/2^{n-1}\ne13$ as $n\le2^{n-1}$ for $n\in\Bbb Z$.

Another reason to believe that $13$ is the sum of the binomial coefficients only is that the sum of the first and penultimate coefficient is different when we consider $\left(\frac{x^{4/3}}2+\frac1{x^{28/15}}\right)^n$ and $\left(\frac1{x^{28/15}}+\frac{x^{4/3}}2\right)^n$.

So we have $\binom n0+\binom n{n-1}=13$ giving $n=12$. For the constant term we require the powers of $x$ to cancel out, and thus $\color{red}{\frac43}i=\frac{28}{15}(12-i)$ giving $i=7$.