Question

数学问题，化简：(1/√2+√4)+(1/√4+√6)+……+(1/(√2n-2)+(√2n))(n为正整数，且n≠2)

RT,化简：(1/√2+√4)+(1/√4+√6)+……+(1/(√2n-2)+(√2n))(n为正整数，且n≠2)
坐等。。。。。。
跪求答案。。。。。

Answer 1

先看最后一项1/(√2n-2)+(√2n))
分子分母同时乘(√2n)-(√2n-2))
得分子:(√2n)-(√2n-2))
分母:2
所以提出公因式1/2
(1/√2+√4)+(1/√4+√6)+……+(1/(√2n-2)+(√2n))
=(1/2)(√4-√2 +.........+(√2n)-(√2n-2))
=(1/2) (-√2+√2n)

Answer 2

[(√2n-2)+(√2n)][(-√2n-2)+(√2n)]=2n-(2n-2)=2
所以1/(√2n-2)+(√2n)=[(-√2n-2)+(√2n)]/2
所以
(1/√2+√4)+(1/√4+√6)+……+(1/(√2n-2)+(√2n))
=(-√2+√4)/2+(-√4+√6)/2+……+[(-√2n-2)+(√2n)]/2
=(-√2+√2n)/2

Answer 3

(1/√2+√4)+(1/√4+√6)+……+(1/(√2n-2)+(√2n))
=（√2-√4）/[(√2+√4)(√2-√4)]+(√4-√6)/[(√4+√6)(√4-√6)]+
……+(√2n-2-√2n)/[(√2n-2+√2n)(√2n-2-√2n)]
=(√2-√4)/(-2)+(√4-√6)/(-2)+……+(√2n-2-√2n)/(-2)
=(-½)×(√2-√4+√4-√6+……+√2n-2-√2n)
=(-½)×(√2-√2n)

Answer 4

1/(√2n-2)+(√2n)=-(√2n-2+√2n)/2
(1/√2+√4)+(1/√4+√6)+……+(1/(√2n-2)+(√2n))
=(-√2+√4-√4+√6+……-√2n-2+√2n)/2
=(√2n-√2 )/2