√(11...1[2n個1]-22...2[n個2])=33...3[n個3]
當(dāng)n=1時,
左邊=√(11...1[2個1]-22...2[1個2])
=√(11-2)
=3
=右邊
原式成立;
設(shè)n=k時成立,即√(11...1[2k個1]-22...2[k個2])=33...3[k個3]
11...1[2k個1]=33...3[k個3]^2+22...2[k個2]
當(dāng)n=k+1時,
√(11...1[2k+2個1]-22...2[k+1個2])
=√[(11...1[2k個1]*100+11)-(22...2[k個2]*10+2)]
=√{[(33...3[k個3]^2+22...2[k個2])*100+11]-(22...2[k個2]*10+2)}
=√{10*33...3[k個3]^2+(11...1[k個1]*10-11...1[k個1])*10*2+9}
=√{10*33...3[k個3]^2+2*10*11...1[k個1]*(10-1)+9}
=√{10*33...3[k個3]^2+2*10*9*11...1[k個1]+9}
=√{10*33...3[k個3]^2+2*3*10*33...3[k個3]+9}
=√{10*33...3[k個3]^2+2*3*(10*33...3[k個3])+3^2}
=√{(10*33...3[k個3]+3)^2]
=10*33...3[k個3]+3
=33...3[k+1個3]
所以
√(11...1[2n個1]-22...2[n個2])=33...3[n個3]成立.
證畢.