11.2. A sequence of real numbers $a_{n}, n=1,2,3, \ldots$ is such that $a_{n+1}=a_{n}+\sqrt{a_{n}+a_{n+1}}, n=1,2,3, \ldots$ and $a_{1}=1$. Find an explicit formula expressing the number $a_{n}$ in terms of $n$.
Answer. $a_{n}=1+2+3+\ldots+n=\frac{n(n+1)}{2}$.

Solution. Let's calculate the first terms of the sequence $a_{n}, n=2,3$. For $a_{2}$, we have by the condition $a_{2}=1+\sqrt{1+a_{2}}$, which means $a_{2}^{2}-3 a_{2}=0$ and $a_{2}>1$, hence $a_{2}=3=1+2$. For $a_{3}$, we then have $a_{3}=3+\sqrt{3+a_{3}}$, which means $a_{3}^{2}-7 a_{3}+6=0$ and $a_{3}>3$, hence $a_{3}=6=1+2+3$. From this, it is easy to guess the general formula $a_{n}=1+2+3+\ldots+n=\frac{n(n+1)}{2}$.

Let's prove it by induction. The base case is already proven for $n=1,2,3$. Suppose the formula is true for $n$, then for $a_{n+1}$ we have by the condition $a_{n+1}=\frac{n(n+1)}{2}+\sqrt{\frac{n(n+1)}{2}+a_{n+1}}$, which means $a_{n+1}^{2}-\left(n^{2}+n+1\right) a_{n+1}+\frac{n^{4}+2 n^{3}-n^{2}-2 n}{4}=0$ and $a_{n+1}>\frac{n(n+1)}{2}$. The discriminant of this quadratic equation is $\left(n^{2}+n+1\right)^{2}-\left(n^{4}+2 n^{3}-n^{2}-2 n\right)=4 n^{2}+4 n+1=(2 n+1)^{2}$, from which $a_{n+1}=\frac{\left(n^{2}+n+1\right) \pm(2 n+1)}{2}=\frac{n^{2}+3 n+2}{2}, \frac{n^{2}-n}{2}$. Taking into account the restriction $a_{n+1}>\frac{n(n+1)}{2}$, we get $a_{n+1}=\frac{n^{2}+3 n+2}{2}=\frac{(n+1)(n+2)}{2}$, which is what we needed to prove.

Grading criteria. The formula for $a_{n}$ is correctly guessed and verified for small $n=2,3: 1$ point. The formula for $a_{n}$ is correctly found and fully justified (with the selection of the correct roots) for small $n=2,3$, but not proven in the general case: 2 points.

There is a proof of the formula in the general case, but no explanation for selecting the correct root: minus 2 points.