√(11...1[2n个1]-22...2[n个2])=33...3[n个3]
当n=1时,
左边=√(11...1[2个1]-22...2[1个2])
=√(11-2)
=3
=右边
原式成立;
设n=k时成立,即√(11...1[2k个1]-22...2[k个2])=33...3[k个3]
11...1[2k个1]=33...3[k个3]^2+22...2[k个2]
当n=k+1时,
√(11...1[2k+2个1]-22...2[k+1个2])
=√[(11...1[2k个1]*100+11)-(22...2[k个2]*10+2)]
=√{[(33...3[k个3]^2+22...2[k个2])*100+11]-(22...2[k个2]*10+2)}
=√{10*33...3[k个3]^2+(11...1[k个1]*10-11...1[k个1])*10*2+9}
=√{10*33...3[k个3]^2+2*10*11...1[k个1]*(10-1)+9}
=√{10*33...3[k个3]^2+2*10*9*11...1[k个1]+9}
=√{10*33...3[k个3]^2+2*3*10*33...3[k个3]+9}
=√{10*33...3[k个3]^2+2*3*(10*33...3[k个3])+3^2}
=√{(10*33...3[k个3]+3)^2]
=10*33...3[k个3]+3
=33...3[k+1个3]
所以
√(11...1[2n个1]-22...2[n个2])=33...3[n个3]成立.
证毕.