1/1*3*5+1/3*5*7+1/5*7*9+1/7*9*11+1/9*11*13+1/11*13*15=
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解决时间 2021-01-31 17:30
- 提问者网友：轻浮
- 2021-01-30 19:33
1/1*3*5+1/3*5*7+1/5*7*9+1/7*9*11+1/9*11*13+1/11*13*15=
最佳答案
- 五星知识达人网友：躲不过心动
- 2021-01-30 20:58
设1/[(2n-1)(2n+1)(2n+3)]=[a/(2n-1)(2n+1)]+[b/(2n+1)(2n+3)]
=[a(2n+3)+b(2n-1)]/[(2n-1)(2n+1)(2n+3)]
=[2(a+b)n+(3a-2b)]/[(2n-1)(2n+1)(2n+3)]
所以：a+b=0，3a-2b=1
联立解得：a=1/5，b=-1/5
所以：1/[(2n-1)(2n+1)(2n+3)]=(1/5)*[1/(2n-1)(2n+1)-1/(2n+1)(2n+3)]
=(1/5){(1/2)[1/(2n-1)-1/(2n+1)]-(1/2)[1/(2n+1)-1/(2n+3)]}
=(1/10)*{[1/(2n-1)-1/(2n+1)]-[1/(2n+1)-1/(2n+3)]}
所以，原式=(1/10)*{[1-(1/3)]-[(1/3)-(1/5)]+[(1/3)-(1/5)]-[(1/5)-(1/7)]+……+[(1/11)-(1/13)]-[(1/13)-(1/15)]}
=(1/10)*{[1-(1/3)]-[(1/13)-(1/15)]}
=64/975
=[a(2n+3)+b(2n-1)]/[(2n-1)(2n+1)(2n+3)]
=[2(a+b)n+(3a-2b)]/[(2n-1)(2n+1)(2n+3)]
所以：a+b=0，3a-2b=1
联立解得：a=1/5，b=-1/5
所以：1/[(2n-1)(2n+1)(2n+3)]=(1/5)*[1/(2n-1)(2n+1)-1/(2n+1)(2n+3)]
=(1/5){(1/2)[1/(2n-1)-1/(2n+1)]-(1/2)[1/(2n+1)-1/(2n+3)]}
=(1/10)*{[1/(2n-1)-1/(2n+1)]-[1/(2n+1)-1/(2n+3)]}
所以，原式=(1/10)*{[1-(1/3)]-[(1/3)-(1/5)]+[(1/3)-(1/5)]-[(1/5)-(1/7)]+……+[(1/11)-(1/13)]-[(1/13)-(1/15)]}
=(1/10)*{[1-(1/3)]-[(1/13)-(1/15)]}
=64/975
全部回答
- 1楼网友：慢性怪人
- 2021-01-30 22:15
裂项法: 4/((2k+1)(2k+3)(2k+5))
= ((2k+5)-(2k+1))/((2k+1)(2k+3)(2k+5))
= 1/((2k+1)(2k+3))-1/((2k+3)(2k+5)).
因此4/(1·3·5)+4/(3·5·7)+...+4/(11·13·15)
= (1/(1·3)-1/(3·5))+(1/(3·5)-1/(5·7))+...+(1/(11·13)-1/(13·15))
= 1/(1·3)-1/(13·15)
= 64/195.
故1/(1·3·5)+1/(3·5·7)+...+1/(11·13·15) = 16/195.
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