依题意可知,a 1+b 1=9,a 2+b 2=9,a 3+b 3=9…,
且a 1+a 2+…+a 10=b 1+b 2+…+b 10=45,
∴(a 1 2+a 2 2+…+a 10 2)-(b 1 2+b 2 2+…b 10 2)=(a 1 2-b 1 2)+(a 2 2-b 2 2)+…+(a 10 2-b 10 2)
=(a 1+b 1)(a 1-b 1)+(a 2+b 2)(a 2-b 2)+…+(a 10+b 10)(a 10-b 10)
=9[(a 1+a 2+…+a 10)-(b 1+b 2+…+b 10)]
=0,
∴a 1 2+a 2 2+…+a 10 2=b 1 2+b 2 2+…b 10 2.