(0)

\[ f_{n}=n\sqrt{1+f_{n+1}} \] とおくと、
\begin{align*} f_{1} & =1\sqrt{1+f_{2}}\\ & =1\sqrt{1+2\sqrt{1+f_{3}}}\\ & =1\sqrt{1+2\sqrt{1+3\sqrt{1+f_{4}}}}\\ & =1\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}} \end{align*} となるので\(f_{1}\)が求める値となる。
両辺を2乗すると、
\[ f_{n}^{\;2}=n^{2}\left(1+f_{n+1}\right) \] \(f_{n}\)を\(n\)の級数と仮定し、\(m\)次までで終わるとすると、
\[ f_{n}=\sum_{k=0}^{m}a_{k}n^{k} \] \(m\)の値は最高次数より、
\[ m^{2}=2+m \] より、
\begin{align*} 0 & =m^{2}-m-2\\ & =\left(m+1\right)\left(m-2\right)\\ m & =-1,2 \end{align*} これより、\(m\)は2となるので、
\[ f_{n}=\sum_{k=0}^{2}a_{k}n^{k} \] \[ a_{2}\ne0 \] となる。
\begin{align*} 0 & =f_{n}^{\;2}-n^{2}\left(1+f_{n+1}\right)\\ & =\left(\sum_{k=0}^{2}a_{k}n^{k}\right)^{2}-n^{2}\left(1+\sum_{k=0}^{2}a_{k}\left(n+1\right)^{k}\right)\\ & =\sum_{k=0}^{4}\sum_{j=0}^{k}a_{k-j}a_{j}n^{k}-n^{2}\left(1+\sum_{k=0}^{2}\sum_{j=0}^{k}a_{k}C\left(k,j\right)n^{j}\right)\\ & =\sum_{k=0}^{4}\sum_{j=0}^{k}a_{k-j}a_{j}n^{k}-n^{2}\left(1+\sum_{j=0}^{2}\sum_{k=j}^{2}a_{k}C\left(k,j\right)n^{j}\right)\\ & =\sum_{k=0}^{4}\sum_{j=0}^{k}a_{k-j}a_{j}n^{k}-n^{2}\left(1+\sum_{k=0}^{2}\sum_{j=k}^{2}a_{j}C\left(j,k\right)n^{k}\right)\\ & =\sum_{k=0}^{4}\sum_{j=0}^{k}a_{k-j}a_{j}n^{k}-\left(n^{2}+\sum_{k=0}^{2}\sum_{j=k}^{2}a_{j}C\left(j,k\right)n^{k+2}\right)\\ & =\sum_{k=0}^{4}\sum_{j=0}^{k}a_{k-j}a_{j}n^{k}-\left(n^{2}+\sum_{k=2}^{4}\sum_{j=k-2}^{2}a_{j}C\left(j,k-2\right)n^{k}\right)\\ & =a_{0}^{2}+n\sum_{j=0}^{1}a_{1-j}a_{j}+n^{2}\left(-1+\sum_{j=0}^{2}\left(a_{2-j}a_{j}-a_{j}\right)\right)+\sum_{k=3}^{4}n^{k}\left(\sum_{j=0}^{k}a_{k-j}a_{j}-\sum_{j=k-2}^{2}a_{j}C\left(j,k-2\right)\right) \end{align*} \(n\)の0次の項より、
\[ a_{0}=0 \] \(n\)の1次の項より、
\begin{align*} 0 & =\sum_{j=0}^{1}a_{1-j}a_{j}\\ & =0 \end{align*} \(n\)の2次の項より、
\begin{align*} 0 & =-1+\sum_{j=0}^{2}\left(a_{2-j}a_{j}-a_{j}\right)\\ & =-1+a_{1}^{\;2}-\sum_{j=1}^{2}a_{j} \end{align*} なので、
\[ a_{2}=a_{1}^{\;2}-a_{1}-1 \] \(n\)の3次の項より、
\begin{align*} 0 & =\sum_{j=0}^{3}a_{k-j}a_{j}-\sum_{j=1}^{2}a_{j}C\left(j,1\right)\\ & =\sum_{j=1}^{2}a_{3-j}a_{j}-\sum_{j=1}^{2}a_{j}j\\ & =2a_{1}a_{2}-\sum_{j=1}^{2}a_{j}j\\ & =2a_{1}a_{2}-a_{1}-2a_{2} \end{align*} \(n\)の4次の項より、
\begin{align*} 0 & =\sum_{j=0}^{4}a_{4-j}a_{j}-\sum_{j=2}^{2}a_{j}C\left(j,2\right)\\ & =\sum_{j=1}^{3}a_{4-j}a_{j}-a_{2}\\ & =a_{1}a_{3}+a_{2}^{\;2}-a_{2}\\ & =a_{2}^{\;2}-a_{2}\\ & =a_{2}\left(a_{2}-1\right) \end{align*} \[ a_{2}=0,1 \] まとめると、
\[ \begin{cases} a_{0}=0\\ a_{2}=a_{1}^{\;2}-a_{1}-1\\ 2a_{1}a_{2}-a_{1}-2a_{2}=0\\ a_{2}=0,1 \end{cases} \] \(a_{2}\ne0\)なので\(a_{2}=0\)は不適。
\(a_{0}=0\;,\;a_{1}=2\;,\;a_{2}=1\)となり、
\begin{align*} f_{n} & =2n+n^{2}\\ & =n\left(n+2\right) \end{align*} となるので
\begin{align*} f_{1} & =1\left(1+2\right)\\ & =3 \end{align*}

(0)-2

天下り的証明
\begin{align*} 3 & =\sqrt{1+8}\\ & =\sqrt{1+2\sqrt{1+15}}\\ & =1\sqrt{1+2\sqrt{1+3\sqrt{1+24}}}\\ & =1\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+35}}}} \end{align*}