√(1+1/1^2+1/2^2)+√(1+1/2^2+1/3^2)+√(1+1/3^2+1/4^2)……√（1+1/2009^2+1/2010^2）
√(1+1/1^2+1/2^2)= √(1+1+1/4)= √9/4=3/2=1+1/2
√(1+1/2^2+1/3^2)= √(1+1/4+1/9)= √49/36=7/6=1+1/(2*3)
√(1+1/3^2+1/4^2)=1+1/3*4
……
√（1+1/2009^2+1/2010^2）=1+1/2009*2010
設√（1+1/a^2+1/(a+1)^2）=1+1/a(a+1)
右邊平方得：[a(a+1)+1]^2/[a(a+1)]^2=[a(a+1)^2+2a(a+1)+1}/[a(a+1)]^2=1+(2a^2+2a+1)/ [a(a+1)]^2=1+[(a^2+2a+1)+a^2]/[a(a+1)]^2=1+[(a+1)^2+a^2] /[a(a+1)]^2=1+1/a^2+1/(a+1)^2與根號下相同所以等式成立.
1/a-1/(a+1)=(a+1-a)/a(a+1)=1/a(a+1)
所以1/a(a+1)=1/a-1/(a+1)成立
原式=1+1/1*2+1+1/2*3+……+1+1/2009*2010=2009+1-1/2+1/2-1/3+1/3-1/4+……+1/2009-1/2010=2009+1-1/2010=2010-1/2010=2009/2010