计算:1/1x3+1/3x5+1/5x7+.+1/31x33+1/33x35,
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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
再问: 可以简便些吗,讲解
再答: 因为 1/1 - 1/3 = 3/3 - 1/3 = 2/3 = 2×1/3 所以,1/(1×3) = 1/2 × (1/1 - 1/3) 同理,1/3 - 1/5 = 5/15 - 3/15 = 2/15 = 2 × 1/15 所以,1/(3×5) = 1/2×(1/3 - 1/5) ………… 1/(33×35) = 1/2×(1/33 - 1/35) 代到公式中,你会发现放多分数都会正、负抵消,只剩下 1 和 -1/35 所以得到最终结果。
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
再问: 可以简便些吗,讲解
再答: 因为 1/1 - 1/3 = 3/3 - 1/3 = 2/3 = 2×1/3 所以,1/(1×3) = 1/2 × (1/1 - 1/3) 同理,1/3 - 1/5 = 5/15 - 3/15 = 2/15 = 2 × 1/15 所以,1/(3×5) = 1/2×(1/3 - 1/5) ………… 1/(33×35) = 1/2×(1/33 - 1/35) 代到公式中,你会发现放多分数都会正、负抵消,只剩下 1 和 -1/35 所以得到最终结果。
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