计算：1/1x3+1/3x5+1/5x7+.+1/31x33+1/33x35,

计算：1/1x3+1/3x5+1/5x7+.+1/31x33+1/33x35,
计算：1/1x3+1/3x5+1/5x7+.+1/31x33+1/33x35,

计算：1/1x3+1/3x5+1/5x7+.+1/31x33+1/33x35,
因为 an=1/[(2n-1)*(2n+1)] = 1/2 [1/(2n-1) - 1/(2n+1)]
所以,
1/1x3+1/3x5+1/5x7+.+1/31x33+1/33x35
= 1/2*[(1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + …… + (1/33 - 1/35)]
= 1/2*(1/1 - 1/35)
=1/2* 34/35
=17/35

1/1x3=½×﹙1-1/3),
1/3x5=½×(1/3-1/5)
1/1x3+1/3x5+1/5x7+......+1/31x33+1/33x35
=½×﹙1-1/3+1/3-1/5+1/5-1/7+……+1/33-1/35)
=½×(1-1/35)
=½×34/35
=17/35