3. If the sequence $\left\{a_{n}\right\}$ is defined by $a_{1}=1, 4 a_{n} a_{n+1}=\left(a_{n}+a_{n+1}-1\right)^{2}, a_{n}>a_{n-1}$, then $a_{n}$ equals
A. $n^{2}$
B. $n$
C. $n(n-1)$
D. $n^{3}$
3. A transforms the given recurrence relation into $a_{n+1}^{2}-2\left(a_{n}+1\right) a_{n+1}+\left(a_{n}-1\right)^{2}=0$.

By substituting $(n-1)$ for the subscript $(n+1)$, and rearranging, we get
$$
a_{n-1}^{2}-2\left(a_{n}+1\right) a_{n-1}+\left(a_{n}-1\right)^{2}=0 \text {. }
$$

From (1) and (2), we know that $a_{n+1}$ and $a_{n-1}$ are the two roots of the equation $x^{2}-2\left(a_{n}+1\right) x+\left(a_{n}-1\right)^{2}=0$. 
Thus, $a_{n+1}+a_{n-1}=2 a_{n}+2$.
Let $b_{n}=a_{n+1}-a_{n}$, then equation (3) can be rewritten as $b_{n}=b_{n-1}+2$, indicating that $\left\{b_{n}\right\}$ is an arithmetic sequence. It is easy to derive that $b_{n}=2 n+1$, hence $a_{n+1}-a_{n}=2 n+1$.
Therefore, $a_{n}=a_{1}+\sum_{i=1}^{n-1}(2 i+1)=1+2 \cdot \frac{n(n-1)}{2}+n-1=n^{2}$. So the answer is A.