北京大学 2022年强基第1题

【解析】令 $\displaystyle n=1 . a_{1}=\frac{a_{1}^{2}+1}{2 a_{1}} a_{1}^{2}=1 \quad a_{1}= \pm 1$ ． $$ \begin{aligned} & \qquad S_{n}=\frac{a_{n}^{2}+n^{2}}{2 a_{n}}=\frac{a_{n}}{2}+\frac{n^{2}}{2 a_{n}} \\ & \qquad S_{n-1}=\frac{a_{n-1}^{2}+(n-1)^{2}}{2 a_{n-1}}=\frac{a_{n-1}}{2}+\frac{(n-1)^{2}}{2 a_{n-1}} \\ & \text { 两式相减 } \frac{a_{n}}{2}-\frac{n^{2}}{2 a_{n}}=-\frac{a_{n-1}}{2}-\frac{(n-1)^{2}}{2 a_{n-1}} \\ & \left(a_{n}-\frac{n^{2}}{a_{n}}\right)^{2}=\left(a_{n-1}+\frac{(n-1)^{2}}{a_{n-1}}\right)^{2} . \\ & a_{n}^{2}+\frac{n^{4}}{a_{n}^{2}}-2 n^{2}=a_{n-1}^{2}+\frac{(n-1)^{4}}{a_{n-1}^{2}}+2(n-1)^{2} \\ & \text { 令 } b_{n}=a_{n}^{2}+\frac{n^{4}}{a_{n}^{2}} \\ & b_{n}=b_{n-1}+2\left((n-1)^{2}+n^{2}\right) . \\ & b_{1}=a_{1}^{2}+\frac{1}{a_{1}^{2}}=2 . \\ & b_{n}=\left(b_{n}-b_{n-1}\right)+\left(b_{n-1}+b_{n-2}\right)+\ldots+\left(b_{2}-b_{1}\right)+b_{1} \\ & =2\left(n^{2}+(n-1)^{2}+\ldots+1+(n-1)^{2}+\ldots+1+0\right) . \\ & =2 \times\left(\frac{1}{6} \cdot n(n+1)(2 n+1)+\frac{1}{6}(n-1) n(2 n-1)\right) \\ & =2 \frac{n}{3}\left(2 n^{2}+1\right) \\ & b_{100}=1333400 . \\ & a_{100}{ }^{2}+\frac{10000}{a_{100}^{2}}=1333400 \end{aligned} $$