写出 $\left(x + \frac{5}{x}\right)^n$ 的二项式展开式的通项。 已知在 $\left(x + \frac{5}{x}\right)^n$ 的展开式中存在常数项，证明 $n$ 必须是偶数。 求 $\left(x + \frac{5}{x}\right)^{10}$ 的展开式中 $x^6$ 的项。 由此或用其他方法，求 $a$ 的值，已知在 $(ax^2 - 1)\left(x + \frac{5}{x}\right)^{10}$ 的展开式中， $x^6$ 的系数为 43875。

1. Write down the general term formula for the given binomial

The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is $\binom{n}{r} a^{n-r} b^r$.
For $\left(x + \frac{5}{x}\right)^n$, we have:
$T_{r+1} = \binom{n}{r} x^{n-r} \left(\frac{5}{x}\right)^r$

2. Simplify the general term

$T_{r+1} = \binom{n}{r} x^{n-r} \frac{5^r}{x^r}$
$T_{r+1} = \binom{n}{r} 5^r x^{n-2r}$

1. Equate the power of $x$ in the general term to 0 for a constant term

From part (i), the general term is $T_{r+1} = \binom{n}{r} x^{n-r} \left(\frac{5}{x}\right)^r = \binom{n}{r} 5^r x^{n-2r}$.
For a constant term, the power of $x$ must be zero:
$n - 2r = 0$

2. Deduce that $n$ must be even

$n = 2r$
Since $r$ is an integer ($0 \leqslant r \leqslant n$), $2r$ is an even number. Therefore, $n$ must be an even number.

1. Identify the terms from the binomial expansion needed to form $x^6$ in the product

The term in $x^6$ in the expansion of $(ax^2 - 1)\left(x + \frac{5}{x}\right)^{10}$ is obtained by multiplying $ax^2$ by the term in $x^4$ from $\left(x + \frac{5}{x}\right)^{10}$, and $-1$ by the term in $x^6$ from $\left(x + \frac{5}{x}\right)^{10}$.

2. Find the term in $x^4$ and the term in $x^6$ from the binomial expansion

The general term is $T_{r+1} = \binom{10}{r} x^{10-r} \left(\frac{5}{x}\right)^r = \binom{10}{r} 5^r x^{10-2r}$.

For the term in $x^6$ (from part iii):
$10 - 2r = 6 \implies r = 2$
Term in $x^6 = \binom{10}{2} 5^2 x^6 = 45(25)x^6 = 1125x^6$

For the term in $x^4$:
$10 - 2r = 4 \implies r = 3$
Term in $x^4 = \binom{10}{3} 5^3 x^4 = 120(125)x^4 = 15000x^4$

3. Form an equation using the given coefficient and solve for $a$

Coefficient of $x^6$ in the product:
$a(15000) - 1(1125) = 43875$
$15000a - 1125 = 43875$
$15000a = 45000$
$a = 3$