等差数列an,求bn=1/[(根号an+1)+(根号an)前n项和,看不懂的有图0 0求过程详解0 0最好不要跳步太严重.因为我很笨肯定看不懂Orz
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等差数列an,求bn=1/[(根号an+1)+(根号an)前n项和,看不懂的有图0 0求过程详解0 0最好不要跳步太严重.因为我很笨肯定看不懂Orz
等差数列an,求bn=1/[(根号an+1)+(根号an)前n项和,看不懂的有图0 0
求过程详解0 0最好不要跳步太严重.因为我很笨肯定看不懂Orz
等差数列an,求bn=1/[(根号an+1)+(根号an)前n项和,看不懂的有图0 0求过程详解0 0最好不要跳步太严重.因为我很笨肯定看不懂Orz
设等差数列an的公差为d,则
bn=1/[(根号an+1)+(根号an)]
=[(根号an+1)-(根号an)]/d
Sn=1/d(根号a2-根号a1+根号a3-根号a2+…+根号an-根号an-1+根号an+1-根号an)
=1/d(根号an+1-根号a1)
设等差数列an的公差为d,则
bn=1/[(根号an+1)+(根号an)]
=[(根号an+1)-(根号an)]/d
Sn=1/d(根号a2-根号a1+根号a3-根号a2+…+根号an-根号an-1+根号an+1-根号an)
=1/d(根号an+1-根号a1)
Sn=n/(an+1-a1)/(根号an+1-根号a1)
Sn=n/(根号an+1+根号a1)
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