设S1=1+1/12+1/22,S2=1+1/22+1/32 ...Sn=1+1/n2+1/(n+1)2,S=√S1+√
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设S1=1+1/12+1/22,S2=1+1/22+1/32 ...Sn=1+1/n2+1/(n+1)2,S=√S1+√S2+…+√Sn ,那么S=
设S1=1+1/12+1/22,S2=1+1/22+1/32...Sn=1+1/n2+1/(n+1)2,S=√S1+√S2+…+√Sn,那么S=
设S1=1+1/12+1/22,S2=1+1/22+1/32...Sn=1+1/n2+1/(n+1)2,S=√S1+√S2+…+√Sn,那么S=
![设S1=1+1/12+1/22,S2=1+1/22+1/32 ...Sn=1+1/n2+1/(n+1)2,S=√S1+√](/uploads/image/z/17760531-3-1.jpg?t=%E8%AE%BES1%3D1%2B1%2F12%2B1%2F22%2CS2%3D1%2B1%2F22%2B1%2F32+...Sn%3D1%2B1%2Fn2%2B1%2F%28n%2B1%292%2CS%3D%E2%88%9AS1%2B%E2%88%9A)
你可以先试试数学归纳法,直接法我先想想,晚上上图
再问: 谢谢,我已经解出了,答案是n加1分之n平方+2n,对了提问数字后面的2是平方,打错了
再答:
有一个小疏忽,最后答案后面再加个n
再问: 谢谢,我已经解出了,答案是n加1分之n平方+2n,对了提问数字后面的2是平方,打错了
再答:
![](http://img.wesiedu.com/upload/e/e9/ee9735466647492a8b85bb5b5b8d00c6.jpg)
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