数列 $u_1,u_2,u_3,\dots$ 满足：对于所有 $n \ge 1$，都有 $u_1 = \dfrac{3}{4}$ 且 $u_{n+1} = u_n + \dfrac{3}{4}\left(\dfrac{1}{2}\right)^{2n}$。

1. Calculate $u_2$ by substituting $n=1$ into the recurrence relation

$u_2 = u_1 + \frac{3}{4}\left(\frac{1}{2}\right)^{2(1)} = \frac{3}{4} + \frac{3}{4}\left(\frac{1}{4}\right) = \frac{3}{4} + \frac{3}{16} = \frac{15}{16}$

2. Calculate $u_3$ by substituting $n=2$ into the recurrence relation

$u_3 = u_2 + \frac{3}{4}\left(\frac{1}{2}\right)^{2(2)} = \frac{15}{16} + \frac{3}{4}\left(\frac{1}{16}\right) = \frac{15}{16} + \frac{3}{64} = \frac{63}{64}$

3. Calculate $u_4$ by substituting $n=3$ into the recurrence relation

$u_4 = u_3 + \frac{3}{4}\left(\frac{1}{2}\right)^{2(3)} = \frac{63}{64} + \frac{3}{4}\left(\frac{1}{64}\right) = \frac{63}{64} + \frac{3}{256} = \frac{255}{256}$

1. Determine the sequence $1 - u_n$ using values from part (i) to form a conjecture.

From part (i), we evaluate $u_2, u_3, u_4$:
$u_1 = \dfrac{3}{4}$
$u_2 = \dfrac{3}{4} + \dfrac{3}{4}\left(\dfrac{1}{4}\right)^1 = \dfrac{15}{16}$
$u_3 = \dfrac{15}{16} + \dfrac{3}{4}\left(\dfrac{1}{4}\right)^2 = \dfrac{63}{64}$
$u_4 = \dfrac{63}{64} + \dfrac{3}{4}\left(\dfrac{1}{4}\right)^3 = \dfrac{255}{256}$

Considering $1 - u_n$:
$1 - u_1 = 1 - \dfrac{3}{4} = \dfrac{1}{4}$
$1 - u_2 = 1 - \dfrac{15}{16} = \dfrac{1}{16} = \left(\dfrac{1}{4}\right)^2$
$1 - u_3 = 1 - \dfrac{63}{64} = \dfrac{1}{64} = \left(\dfrac{1}{4}\right)^3$
$1 - u_4 = 1 - \dfrac{255}{256} = \dfrac{1}{256} = \left(\dfrac{1}{4}\right)^4$

By observing the pattern, the conjecture is:
$1 - u_n = \left(\dfrac{1}{4}\right)^n$
Thus, $u_n = 1 - \left(\dfrac{1}{4}\right)^n$

2. State the proposition and prove the base case $n = 1$.

Let $P_n$ be the statement that $u_n = 1 - \left(\dfrac{1}{4}\right)^n$ for all $n \in \mathbb{Z}^+$.

When $n = 1$:
$\text{LHS} = u_1 = \dfrac{3}{4}$ (given)
$\text{RHS} = 1 - \left(\dfrac{1}{4}\right)^1 = \dfrac{3}{4}$

Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

3. Show that if the proposition is true for $n=k$, it is also true for $n=k+1$.

Assume $P_k$ is true for some $k \in \mathbb{Z}^+$, i.e., $u_k = 1 - \left(\dfrac{1}{4}\right)^k$.
We want to show that $P_{k+1}$ is true, i.e., $u_{k+1} = 1 - \left(\dfrac{1}{4}\right)^{k+1}$.

Using the given recurrence relation:
$u_{k+1} = u_k + \dfrac{3}{4}\left(\dfrac{1}{2}\right)^{2k}$
$= \left[ 1 - \left(\dfrac{1}{4}\right)^k \right] + \dfrac{3}{4}\left(\dfrac{1}{4}\right)^k$
$= 1 - \left(\dfrac{1}{4}\right)^k + \dfrac{3}{4}\left(\dfrac{1}{4}\right)^k$
$= 1 - \dfrac{1}{4}\left(\dfrac{1}{4}\right)^k$
$= 1 - \left(\dfrac{1}{4}\right)^{k+1}$

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

4. Write the concluding statement for mathematical induction.

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

1. Use the recurrence relation from the question to express the term being summed as a difference.

From the given recurrence relation, $u_{r+1} = u_r + \frac{3}{4}\left(\frac{1}{2}\right)^{2r}$.
Rearranging gives:
$u_{r+1} - u_r = \frac{3}{4}\left(\frac{1}{2}\right)^{2r}$
Therefore, the summation becomes:
$\sum_{r=2}^{N} \frac{3}{4}\left(\frac{1}{2}\right)^{2r} = \sum_{r=2}^{N} (u_{r+1} - u_r)$

2. Apply the method of differences and substitute the general formula for $u_n$ from part (ii).

$\sum_{r=2}^{N} (u_{r+1} - u_r) = (u_3 - u_2) + (u_4 - u_3) + \dots + (u_{N+1} - u_N)$
$= u_{N+1} - u_2$
From part (ii), we established the conjecture $u_n = 1 - \left(\frac{1}{4}\right)^n$. Substituting this into our expression:
$= \left[ 1 - \left(\frac{1}{4}\right)^{N+1} \right] - \left[ 1 - \left(\frac{1}{4}\right)^2 \right]$
$= \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^{N+1}$
$= \frac{1}{16} - \left(\frac{1}{4}\right)^{N+1}$

1. Concisely evaluate the sum from part (iii) using the results of earlier parts.

From part (ii), we established $u_n = 1 - \left(\frac{1}{4}\right)^n$.
Using the recurrence relation $u_{r+1} - u_r = \frac{3}{4}\left(\frac{1}{2}\right)^{2r}$, the sum in part (iii) is evaluated by the method of differences:
$\sum_{r=2}^{N} \frac{3}{4}\left(\frac{1}{2}\right)^{2r} = \sum_{r=2}^{N} (u_{r+1} - u_r) = u_{N+1} - u_2$
$= \left[1 - \left(\frac{1}{4}\right)^{N+1}\right] - \left[1 - \left(\frac{1}{4}\right)^2\right] = \frac{1}{16} - \left(\frac{1}{4}\right)^{N+1}$

2. Set up the inequality and simplify to isolate the exponential term.

We require the sum to exceed $\frac{3}{50}$:
$\frac{1}{16} - \left(\frac{1}{4}\right)^{N+1} > \frac{3}{50}$
$\left(\frac{1}{4}\right)^{N+1} < \frac{1}{16} - \frac{3}{50}$
$\left(\frac{1}{4}\right)^{N+1} < \frac{25 - 24}{400}$
$\left(\frac{1}{4}\right)^{N+1} < \frac{1}{400}$
$4^{N+1} > 400$

3. Solve the exponential inequality using logarithms.

Taking the natural logarithm on both sides:
$(N+1)\ln 4 > \ln 400$
$N+1 > \frac{\ln 400}{\ln 4}$
$N > 4.3219 - 1$
$N > 3.3219$
Since $N$ must be an integer, the smallest value is $N = 4$.