1除以(X+1)(X+3)加上1除以(X+3)(X+5)加上1除以(X+5)(X+7)加上1除以(X+7)(X+9)加上

1除以(X+1)(X+3)加上1除以(X+3)(X+5)加上1除以(X+5)(X+7)加上1除以(X+7)(X+9)加上1除以(X+9)(X+11)..
求上述问题的算法与规律!
熊猫妹 1年前 已收到3个回答 举报

草包龙 幼苗

共回答了24个问题采纳率:83.3% 举报

1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+……+1÷(x+2n-1)(x+2n+1)
=1/(x+1)(x+3)+1/(x+3)(x+5)+1/(x+5)(x+7)+1/(x+7)(x+9)+……+1/(x+2n-1)(x+2n+1)
=[1/(x+1)-1/(x+3)]/2+[1/(x+3)-1/(x+5)]/2+[1/(x+5)-1/(x+7)]/2+……+[1/(x+2n-1)-1/(x+2n+1)]/2
=1/[2(x+1)]-1/[2(x+3)]+1/[2(x+3)]-1/[2(x+5)]+1/[2(x+5)]-1/[2(x+7)]+……+1/[2(x+2n-1)]-1/[2(x+2n+1)]
=1/[2(x+1)]-1/[2(x+2n+1)]
=(x+2n+1-x-1)/[2(x+1)(x+2n+1)]
=n/[(x+1)(x+2n+1)]
楼主追问“我不是他舅”先生:一直加到101,结果是?
楼主这里说的101,应该是一直加到1÷/(x+101)(x+103)吧?
套用上面我推导出来的公式,应该是n=51
有:
1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+……+1÷(x+101)(x+103)
=51/[(x+1)(x+2×51+1)]
=51/[(x+1)(x+103)]
要是还想继续算的话,就是把分母写成多项式的形式.那么,楼主所求的结果就是:
51/(x^2+104x+103)

1年前 追问

7

熊猫妹 举报

1÷/(X+99)(x+101)

举报 草包龙

1÷/(X+99)(x+101)什么意思?最后一项是1÷/(x+99)(x+101)? 那就是n=50 代入我在前面推导出来的公式,就是: 1÷(x+1)(x+3)+1÷(x+3)(x+5)+1÷(x+5)(x+7)+1÷(x+7)(x+9)+……+1÷(x+99)(x+101) =50/[(x+1)(x+2×50+1)] =50/[(x+1)(x+101)] =50/(x^2+102x+101)

熊猫妹 举报

是2X+102除以(X+1)(X+101)吗?

举报 草包龙

不是。 是:50/[(x+1)(x+101)]

fired1 幼苗

共回答了5个问题 举报

0.5*【1/(X+1)-1/(X+3)】+
0.5*【1/(X+3)-1/(X+5)】+
0.5*【1/(X+5)-1/(X+7)】+
0.5*【1/(X+7)-1/(X+9)】+
0.5*【1/(X+9)-1/(X+11)】+······
=0.5*【1/(X+1)-1/(X+3)+1/(X+3)-1/(X+5)+1/(X+5)-1/(X+7)+····...

1年前

2

马元元 精英

共回答了21805个问题 举报

原式=1/2[1/(X+1)-1/(X+3)+1/(X+3)-1/(X+5)+1/(X+5)-1/(X+7)+1/X+7)-1/(X+9)+1/(X+9)-1/(X+11)]
=1/2[1/(x+1)-1/(x+11)]
=5/(X²+12X+11)

1年前

0
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