Find m, n, p of given expression with some conditions

Question

Find $m, n, p$ such that in the expansion of the expression $ \left( x^m + \frac{1}{x^p} \right)^n$ the 12th and the 24th terms contain $x$, respectively $x^5$, and furthermore, this expansion also contains one free term (without $x$).

By using what we are given, I got the following: $$mn - k(m + p) = 0$$ $$mn - 11(m + p) = 1$$ $$mn - 23(m + p) = 5$$

From these expressions you can easily conclude that $mn = -\frac{8}{3}$ and $m + p = -\frac{1}{3}$. By using these in the first equation, we get $k = 8$.

From this point we have three equations with 3 unknown variables. I have no idea what to do next.

Thank you in advance!

Answer 1

Expanding the binomial we obtain \begin{align*} \left( x^m + \frac{1}{x^p} \right)^n&=\sum_{k=0}^n\binom{n}{k}x^{mk}x^{-p(n-k)}\\ &=\sum_{k=0}^n\binom{n}{k}x^{(m+p)k-pn} \end{align*}

According to the requirements we derive three equations: \begin{align*} 11(m+p)-pn&=1\tag{1}\\ 23(m+p)-pn&=5\tag{2}\\ k(m+p)-pn&=0\tag{3}\\ \end{align*} Subtracting (2) from (1) and (3) from (2) gives \begin{align*} 12(m+p)&=4\\ (k-23)(m+p)&=-5 \end{align*}

From the first we derive $m+p=\frac{1}{3}$ and obtain by putting this value in the second equation according to OPs result \begin{align*} k&=8 \end{align*} With $k=8$ we obtain from (3) \begin{align*} p\cdot n=k(m+p)=\frac{8}{3}\tag{4} \end{align*}

Since we have only to find a proper tripel $(n,m,p)$ we are free to specify appropriate values of $p$ and $n$ in (4). According to the requirement that the exponent of $x$ of the $24$-th term is $5$, we have to set $n$ at least to $24$ and we just take this value for $n$.

We obtain with $n=24$ from (4) and since $m+p=\frac{1}{3}$ \begin{align*} p=\frac{1}{9}, m=\frac{2}{9} \end{align*} We finally get a solution \begin{align*} \left(x^{\frac{2}{9}}+x^{-\frac{1}{9}}\right)^{24}=x^{-\frac{8}{3}}(1+x^{\frac{1}{3}})^{24} \end{align*}