(1/根号2+1+1/根号3+根号2+1/根号4+根号3+…+1/根号2012+根号2011)(根号2012+1),计算

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(1/根号2+1+1/根号3+根号2+1/根号4+根号3+…+1/根号2012+根号2011)(根号2012+1),计算

1/(√2+1)+……+1/(√2012+√2011)
=(√2-1)/[(√2+1)(√2-1)]+(√3-√2)/[(√3+√2)(√3-√2)]+……+(√2012-√2011)/[(√2012+√2011)(√2012-√2011)]
=(√2-1)+(√3-√2)+……+(√2012-√2011)
=√2012-1

看题目应该是还要乘以√2012+1吧,
原式=(√2012-1)(√2012+1)
=2012-1=2011

中间多次用到平方差公式~