2022-M2-05

$\begin{array}{cl}
& (a+x)^n \\
= & a^n+C^n_1a^{n-1}x+C^n_2a^{n-2}x^2 +\ldots +x^n \\
= & a^n +a^{n-1}nx+\dfrac{a^{n-2}n(n-1)}{2}x^2 +\ldots+x^n
\end{array}$

Since $\mu_2=-10$, we have

$\begin{array}{rcl}
\dfrac{a^{n-2}n(n-1)}{2} & = & -10 \\
a^{n-2}n(n-1) & = & -20 \ \ldots \unicode{x2460}
\end{array}$

Note that $n$ is a positive integer greater than $1$. Then $n(n-1)$ is also a positive integer.

Hence, $a^{n-2}$ must be a negative number.

If $n$ is an even number, then $n-2$ is also an even number. $a^{n-2}$ will be a positive number for all values of $a$. Hence, $n$ must be an odd number.

Since $a^{n-2}$ is a negative number and $n-2$ is an odd number, then $a$ must be a negative number.

$\begin{array}{cl}
& (bx-1)^n \\
= & (-1+bx)^n \\
= & (-1)^n+C^n_1(-1)^{n-1}(bx)+C^n_2(-1)^{n-2}(bx)^2+\ldots+(bx)^n \\
= & -1+bnx-\dfrac{b^2n(n-1)}{2}x^2+\ldots+b^nx^n \text{ , $\because n$ is an odd number.}
\end{array}$

Since $\lambda_0=\mu_0$, we have

$\begin{array}{rcl}
\lambda_0 & = & \mu_0 \\
-1 & = & a^n \\
a & = & -1
\end{array}$

Since $\lambda_1 =2\mu_1$, we have

$\begin{array}{rcl}
\lambda_1 & = & 2\mu_1 \\
bn& = & 2a^{n-1}n \\
b & = & 2(-1)^{n-1} \text{ , $\because n$ is an odd number.} \\
b & = & 2
\end{array}$

Sub. $a=-1$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
(-1)^{n-2}n(n-1) & = & -20 \\
n(n-1) & = & 20 \text{ , $\because n$ is an odd number.} \\
n^2-n-20 & = & 0 \\
(n-5)(n+4) & = & 0
\end{array}$

$\therefore n=5$ or $n=-4$ (rejected).