Question

19．已知（$\sqrt{2}$+1）²¹=a+b$\sqrt{2}$，其中a和b為正整數(shù)，則b與27的最大公約數(shù)是1．

Answer 1

分析 （$\sqrt{2}$+1）²¹的展開式的通項公式為：T_r+1=${∁}_{21}^{r}$$（\sqrt{2}）^{r}$，（r∈N，0≤r≤21）．當(dāng)r為奇數(shù)時，b$\sqrt{2}$=${∁}_{21}^{1}×\sqrt{2}$+${∁}_{21}^{3}（\sqrt{2}）^{3}$+…${∁}_{21}^{21}（\sqrt{2}）^{21}$=$\sqrt{2}$$（{∁}_{21}^{1}+2{∁}_{21}^{3}+…+{2}^{10}{∁}_{21}^{21}）$，可得b=${∁}_{21}^{1}$+2${∁}_{21}^{3}$+…+${2}^{10}{∁}_{21}^{21}$=1025×21+514×${∁}_{21}^{3}$+264${∁}_{21}^{5}$+144${∁}_{21}^{7}$+96×${∁}_{21}^{11}$，即可得出．

解答 解：∵（$\sqrt{2}$+1）²¹的展開式的通項公式為：T_r+1=${∁}_{21}^{r}$$（\sqrt{2}）^{r}$，（r∈N，0≤r≤21）．
當(dāng)r為奇數(shù)時，b$\sqrt{2}$=${∁}_{21}^{1}×\sqrt{2}$+${∁}_{21}^{3}（\sqrt{2}）^{3}$+…${∁}_{21}^{21}（\sqrt{2}）^{21}$=$\sqrt{2}$$（{∁}_{21}^{1}+2{∁}_{21}^{3}+…+{2}^{10}{∁}_{21}^{21}）$，
∴b=${∁}_{21}^{1}$+2${∁}_{21}^{3}$+…+${2}^{10}{∁}_{21}^{21}$=1025×21+514×${∁}_{21}^{3}$+264${∁}_{21}^{5}$+144${∁}_{21}^{7}$+96×${∁}_{21}^{11}$與3互質(zhì)，
∴b與27的最大公約數(shù)是1．
故答案為：1．

點評 本題考查了二項式定理的應(yīng)用、互質(zhì)的性質(zhì)，考查了推理能力與計算能力，屬于中檔題．