【初中数学】﹙1-1/2²﹚﹙1-1/3²﹚...﹙1-1/2013²﹚
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分母=2013*2=4026,分子=1+2013=2014,得数=2014/4026
过程
﹙1-1/2²﹚﹙1-1/3²﹚...﹙1-1/2013²﹚
=3/4*﹙1-1/3²﹚...﹙1-1/2013²﹚
=4/6*﹙1-1/4²﹚...﹙1-1/2013²﹚
=5/8*﹙1-1/5²﹚...﹙1-1/2013²﹚
=6/10*﹙1-1/6²﹚...﹙1-1/2013²﹚
=7/12...
找出规律到第一个小括号得数的分母是2的两倍,分子是2+1
找出规律到第二个小括号得数的分母是3的两倍,分子是3+1
···
找出规律到第2012个小括号得数的分母是2013的两倍,分子是2013+1
过程
﹙1-1/2²﹚﹙1-1/3²﹚...﹙1-1/2013²﹚
=3/4*﹙1-1/3²﹚...﹙1-1/2013²﹚
=4/6*﹙1-1/4²﹚...﹙1-1/2013²﹚
=5/8*﹙1-1/5²﹚...﹙1-1/2013²﹚
=6/10*﹙1-1/6²﹚...﹙1-1/2013²﹚
=7/12...
找出规律到第一个小括号得数的分母是2的两倍,分子是2+1
找出规律到第二个小括号得数的分母是3的两倍,分子是3+1
···
找出规律到第2012个小括号得数的分母是2013的两倍,分子是2013+1
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用因式分解做:
原式=(1-1/2)(1+1/2)(1-1/3)(1+1/3)......(1-1/2013^2)(1+1/2013^2)
=(1/2)(3/2)(2/3)(4/3).....(2011/2012)(2013/2012)(2012/2013)(2014/2013)
中间项消除,剩下首尾项
=(1/2)*1*1*1....*1*(2014/2013)
=(1/2)(2014/2013)
=1007/2013
原式=(1-1/2)(1+1/2)(1-1/3)(1+1/3)......(1-1/2013^2)(1+1/2013^2)
=(1/2)(3/2)(2/3)(4/3).....(2011/2012)(2013/2012)(2012/2013)(2014/2013)
中间项消除,剩下首尾项
=(1/2)*1*1*1....*1*(2014/2013)
=(1/2)(2014/2013)
=1007/2013
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原式=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)......(1-1/99)(1+1/99)(1-1/100)(1+1/100)
=(1/2)[(3/2)(2/3)(4/3)(3/4)(5/4).....(98/99)(100/99)(99/100)](101/100)
=(1/2)*1*1*1....*1*(101/100)
=(1/2)(101/100)
=101/200
=(1/2)[(3/2)(2/3)(4/3)(3/4)(5/4).....(98/99)(100/99)(99/100)](101/100)
=(1/2)*1*1*1....*1*(101/100)
=(1/2)(101/100)
=101/200
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先由平方差公式得
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)......(1-1/2013)(1+1/2013)
算出括号的每个数
=(1/2) X (3/2) X( 2/3) X (4/3) X(3/4) X(5/4) X(4/5)x(6/5)x........x (2011/2012)X (2013/2012) (2012/2013)x (2014/2013)
约分得
=(1/2 )X (2014/2013)
=1007/2013
备注:括号是为了让你看清楚些
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)......(1-1/2013)(1+1/2013)
算出括号的每个数
=(1/2) X (3/2) X( 2/3) X (4/3) X(3/4) X(5/4) X(4/5)x(6/5)x........x (2011/2012)X (2013/2012) (2012/2013)x (2014/2013)
约分得
=(1/2 )X (2014/2013)
=1007/2013
备注:括号是为了让你看清楚些
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=﹛1-(1/2)²﹜·﹛1-(1/3)²﹜·﹛1-﹙1/4﹚²﹜······﹛1-﹙1/n²﹚﹜
=﹙1-1/2﹚·﹙1﹢1/2﹚·﹙1-1/3﹚·﹙1﹢1/3﹚·﹙1-1/4﹚·﹙1﹢1/4﹚······﹙1-1/n﹚·﹙1﹢1/n﹚
=1/2×3/2 ×2/3 ×4/3×3/4×········×﹙n-1﹚/n×﹙n﹢1﹚/n
=1/2×﹙n-1﹚/n
=﹙n-1﹚/2n
=﹙1-1/2﹚·﹙1﹢1/2﹚·﹙1-1/3﹚·﹙1﹢1/3﹚·﹙1-1/4﹚·﹙1﹢1/4﹚······﹙1-1/n﹚·﹙1﹢1/n﹚
=1/2×3/2 ×2/3 ×4/3×3/4×········×﹙n-1﹚/n×﹙n﹢1﹚/n
=1/2×﹙n-1﹚/n
=﹙n-1﹚/2n
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原式=[(2²-1)/2²][(3²-1)/3²]......[(2013²-1)/2013²]
=[(2-1)(2+1)(3-1)(3+1)......(2013-1)(2013+1)]/(2²x3²x...x2013²)
=(1x2x3x...x2014)(3x4x...x2012)/(2x3x...2013)²
=2014/(2x2013)
=1007/2013
=[(2-1)(2+1)(3-1)(3+1)......(2013-1)(2013+1)]/(2²x3²x...x2013²)
=(1x2x3x...x2014)(3x4x...x2012)/(2x3x...2013)²
=2014/(2x2013)
=1007/2013
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