14. 设 $\left\{a_{n}\right\}$ 是首项为 1 的正项数列, 且 $(n+1) \cdot a_{n+1}^{2}-n a_{n}^{2}+a_{n+1} a_{n}=0(n=1,2,3, \cdots)$, 则它的通项公式是
$\because(n+1) a_{n+1}^{2}-n a_{n}^{2}+a_{n} a_{n+1}=0$,

$\therefore\left[(n+1) a_{n+1}-n a_{n}\right] \cdot\left(a_{n+1}+a_{n}\right)=0$,

$\because a_{n}>0, \quad \therefore a_{n}+a_{n+1}>0$,

$\therefore(n+1) a_{n+1}-n a_{n}=0$.

方法一 $\frac{a_{n+1}}{a_{n}}=\frac{n}{n+1}$.

$\therefore \frac{a_{2}}{a_{1}} \cdot \frac{a_{3}}{a_{2}} \cdot \frac{a_{4}}{a_{3}} \cdot \frac{a_{5}}{a_{4}} \ldots \ldots \frac{a_{n}}{a_{n-1}}$

$=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \ldots \cdot \frac{n-1}{n}$,

$\therefore \frac{a_{n}}{a_{1}}=\frac{1}{n}$.

又 $\because a_{1}=1, \therefore a_{n}=\frac{1}{n} a_{1}=\frac{1}{n}$.

方法二 $(n+1) a_{n+1}-n a_{n}=0$,

$\therefore n a_{n}=(n-1) a_{n-1}=\cdots=1 \times a_{1}=1$,

$\therefore n a_{n}=1, a_{n}=\frac{1}{n}$.