数列 $u_0,\ u_1,\ u_2,\ \dots$ 满足 $u_0 = 0$ 且 $u_n = u_{n-1} - 2^{-n}\left(n^2 - 4n + 2\right),\quad\text{对于所有 } n \in \mathbb{Z}^+.$

1. Show that the statement is true for the base case $n=1$.

Let $P_n$ be the statement that $u_n = n^2 2^{-n}$.

For $n=1$, using the given recurrence relation:
$u_1 = u_0 - 2^{-1}(1^2 - 4(1) + 2) = 0 - \frac{1}{2}(-1) = \frac{1}{2}$

Using the formula to be proven:
$u_1 = 1^2 (2^{-1}) = \frac{1}{2}$

Since both yield the same value, $P_1$ is true.

2. State the inductive hypothesis.

Assume that $P_k$ is true for some positive integer $k \in \mathbb{Z}^+$. That is,
$u_k = k^2 2^{-k}$

3. Use the inductive hypothesis to show that $P_{k+1}$ is true.

We want to show that $u_{k+1} = (k+1)^2 2^{-(k+1)}$.

Using the given recurrence relation for $n = k+1$:

Hence, $P_{k+1}$ is true whenever $P_k$ is true.

4. Conclude the proof by mathematical induction.

Since $P_1$ is true, and $P_k$ being true implies that $P_{k+1}$ is true, by the Principle of Mathematical Induction, $P_n$ is true for all $n \in \mathbb{Z}^+$.

1. Rearrange the given recurrence relation to match the expression inside the summation.

From the given relation $u_n = u_{n-1} - 2^{-n}\left(n^2 - 4n + 2\right)$, we can rearrange it as:
$u_n - u_{n-1} = -2^{-n}\left(n^2 - 4n + 2\right)$

2. Rewrite the sum using the rearranged expression and apply the method of differences.

$\sum_{n=1}^{N} \left[ -2^{-n}\left(n^2 - 4n + 2\right)\right] = \sum_{n=1}^{N} (u_n - u_{n-1})$
$= (u_1 - u_0) + (u_2 - u_1) + \dots + (u_N - u_{N-1})$
$= u_N - u_0$

3. Substitute the explicitly proven formula for $u_n$ and the initial condition to find the final result.

From part (i), $u_n = n^2 2^{-n}$. We are also given $u_0 = 0$.
$u_N - u_0 = N^2 2^{-N} - 0$
$= N^2 2^{-N}$

1. Evaluate the sum to $N$ terms by expressing it as a difference of consecutive terms using the given recurrence relation.

From the given recurrence relation, we have:
$-2^{-n}\left(n^2 - 4n + 2\right) = u_n - u_{n-1}$
Summing this from $n=1$ to $N$ gives:
$\sum_{n=1}^{N} \left[ -2^{-n}\left(n^2 - 4n + 2\right)\right] = \sum_{n=1}^{N} (u_n - u_{n-1})$
Using the method of differences:
$= (u_1 - u_0) + (u_2 - u_1) + \dots + (u_N - u_{N-1})$
$= u_N - u_0$
Using the expression $u_N = N^2 2^{-N}$ from part (i) and $u_0 = 0$:
$\sum_{n=1}^{N} \left[ -2^{-n}\left(n^2 - 4n + 2\right)\right] = N^2 2^{-N}$

2. Take the limit of the sum to $N$ terms as $N \to \infty$ to state the sum to infinity.

The sum to infinity is the limit of the partial sum as $N \to \infty$:
$S_\infty = \lim_{N \to \infty} \frac{N^2}{2^N}$
Since the exponential term $2^N$ grows significantly faster than the polynomial term $N^2$ as $N \to \infty$, the fraction approaches $0$.
$S_\infty = 0$