(1+1/1*3)*(1+1/2*4)*(1+1/3*5)一直加到(1+1/98*100)等于多少
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![(1+1/1*3)*(1+1/2*4)*(1+1/3*5)一直加到(1+1/98*100)等于多少](/uploads/image/z/942431-23-1.jpg?t=%EF%BC%881%2B1%2F1%2A3%29%2A%281%2B1%2F2%2A4%29%2A%281%2B1%2F3%2A5%29%E4%B8%80%E7%9B%B4%E5%8A%A0%E5%88%B0%281%2B1%2F98%2A100%29%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91)
(1+1/1*3)*(1+1/2*4)*(1+1/3*5)一直加到(1+1/98*100)等于多少
(1+1/1*3)*(1+1/2*4)*(1+1/3*5)一直加到(1+1/98*100)等于多少
(1+1/1*3)*(1+1/2*4)*(1+1/3*5)一直加到(1+1/98*100)等于多少
∵通项an=[1+1/n*(n+2)]=(n+1)2/n*(n+2).∴[1+1/1*3][1+1/2*4][1+1/3*5]...[1+1/97*99][1+1/98*100]=(22/1*3)(32/2*4)(42/3*5).(982/97*99)(992/98*100)注意到分子可以由相邻的两个分母抵消=2*99/100=99/50
通项an=[1+1/n*(n+2)]=(n+1)²/n*(n+2).
原式=(2²/1*3)(3²/2*4)(4²/3*5).....(98²/97*99)(99²/98*100)(100²/99*101)
=2*100/101
=200/101.
该过程要约分。
求采纳为满意回答。