已知实数序列x0x1x2xn的构成规律由递推关系给出

　　由题意知数列的各项都是正数且递增，即x0＜x1＜x2＜…＜xn＜…∵xn2=（xn-1+1 xn?1 ）2=xn?12+1 xn?12 +2，xn?12＝(xn?2+1 xn?2 )2=xn?22+1 xn?22 +2，…x22＝(x1+1 x1 )2=x12+1 x12 +2，x12＝(x0+1 x0 )2=x02+1 x02 +2，上述几个式子相加，得：xn2=1 xn?12 +1 xn?22 +…+1 x12 +1 x02 +x02+2n，①当n=1000时，则①式得：x10002＞x02+2×1000=52+2000=2025，∴x1000＞45．又当n=100时，由①式得：x1002＞x02+2×100=152，则x10002＝1 x9992 +1 x9982 +1 x9972 +1 x9962 +…+1 x12 +1 x02 +x02+2×1000＜1 x1002 +1 x1002 +…+1 x1002 900个 +1 x02 +1 x02 +…+1 x02 100个 +2025≤900 152 +100 52 +2025=4+4+2025＜452+9＜452+2×45×0.1+0.12=（45+0.1）2=45.12，故x1000＜45.1．综上，45＜x1000＜45.1．