求规律值
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解题思路: 其他
解题过程:
解:【1】对任意正整数n(n=1, 2, 3,4, 5, ,,,,,),恒有:
1/[(2n-1)(2n+1)]=(1/2)×{[1/(2n-1)] -[1/(2n+1)]}
【2】由上面结果可知:
1/(1×3)=(1/2)[1-(1/3)]
1/(3×5)=(1/2)[(1/3)-(1/5)]
1/(5×7)=(1/2)[(1/5)-(1/7)]
.........................................
1/(49×51)=(1/2)[(1/49)-(1/51)]
上面式子累加,可得:
原式=(1/2)[1-(1/51)]=25/51
最终答案:略
解题过程:
解:【1】对任意正整数n(n=1, 2, 3,4, 5, ,,,,,),恒有:
1/[(2n-1)(2n+1)]=(1/2)×{[1/(2n-1)] -[1/(2n+1)]}
【2】由上面结果可知:
1/(1×3)=(1/2)[1-(1/3)]
1/(3×5)=(1/2)[(1/3)-(1/5)]
1/(5×7)=(1/2)[(1/5)-(1/7)]
.........................................
1/(49×51)=(1/2)[(1/49)-(1/51)]
上面式子累加,可得:
原式=(1/2)[1-(1/51)]=25/51
最终答案:略