1/1+√2+1/√2+√3+…+1/√2013+√2014=
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1/1+√2+1/√2+√3+…+1/√2013+√2014=1/1+√2+1/√2+√3+…+1/√2013+√2014=1/1+√2+1/√2+√3+…+1/√2013+√2014=答:分母有理化
1/1+√2+1/√2+√3+…+1/√2013+√2014=
1/1+√2+1/√2+√3+…+1/√2013+√2014=
1/1+√2+1/√2+√3+…+1/√2013+√2014=
答:
分母有理化问题:
1/1+√2+1/√2+√3+…+1/√2013+√2014
=(√2-1)/ [(√2+1)(√2-1)] +(√3-√2)/[(√3+√2)(√3-√2)+.+(√2014-√2013) / [(√2014+√2013)(√2014-√2013)
=(√2-1)+(√3-√2)+(√4-√3)+.+(√2014-√2013)
=√2014 -1
1/1+√2+1/√2+√3+…+1/√2013+√2014=
/√1-√2/+/√2-√3/+…+/√99-√100/
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0-1/(√1+√0)-1/(√2+√1)-1/(√3+√2)+……+1/(√100-√99)
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