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1/1x3x5+1/3x5x7+1/5x7x9+1/7x9x11+1/9x11x13+1/11x13x15
题目详情
1/1x3x5+1/3x5x7+1/5x7x9+1/7x9x11+1/9x11x13+1/11x13x15
▼优质解答
答案和解析
分析:1/1x3x5=1/4×(1/1*3 -1/3*5)
1/3x5x7=1/4×(1/3*5 -1/5*7)
1/5x7x9=1/4×(1/5*7 - 1/7*9)
1/7*9*11=1/4×(1/7*9 -1/9*11)
.
1/2003x2005x2007=1/4×(1/2003*2005 -1/2005*207)
所有的等式相加有
1/1x3x5+1/3x5x7+1/5x7x9+.+1/2003x2005x2007
=1/4×(1/1*3 -1/3*5 +1/3*5 -1/5*7+.+1/2003*2005-1/2005*2007)
=1/4×(1/1*3 - 1/2005*2007)
=335336/4024035
结论:1/n(n+1)(n+2)=1/2×[1/n(n+1) - 1/(n+1)(n+2)]
1/n(n+2)(n+4)=1/4×[1/n(n+2) - 1/(n+2)(n+4)]
1/3x5x7=1/4×(1/3*5 -1/5*7)
1/5x7x9=1/4×(1/5*7 - 1/7*9)
1/7*9*11=1/4×(1/7*9 -1/9*11)
.
1/2003x2005x2007=1/4×(1/2003*2005 -1/2005*207)
所有的等式相加有
1/1x3x5+1/3x5x7+1/5x7x9+.+1/2003x2005x2007
=1/4×(1/1*3 -1/3*5 +1/3*5 -1/5*7+.+1/2003*2005-1/2005*2007)
=1/4×(1/1*3 - 1/2005*2007)
=335336/4024035
结论:1/n(n+1)(n+2)=1/2×[1/n(n+1) - 1/(n+1)(n+2)]
1/n(n+2)(n+4)=1/4×[1/n(n+2) - 1/(n+2)(n+4)]
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