lim[1/(1x4)+1/(4x7)+1/(7x10)+...+1/(3n-2)x(3n+1)]=_____

wangshu152022-10-04 11:39:543条回答

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爆爆MM 共回答了21个问题 | 采纳率81%: 1/(1x4)+1/(4x7)+1/(7x10)+...+1/(3n-2)x(3n+1)]
=1/3*(1-1/4+1/4-1/7+.+1/(3n-2)-1/(3n+1))
=1/3*(1-1/(3n+1))
所以
lim[1/(1x4)+1/(4x7)+1/(7x10)+...+1/(3n-2)x(3n+1)]
=lim1/3*(1-1/(3n+1))
=1/3; 1年前

gemshilei 共回答了644个问题 | 采纳率: 分
1/*(1*4)＝1/3×[1－1/4]
1/(4*7)＝1/3×[1/4－1/7]
1/(7*10)＝1/3×[1/7－1/10]
.......
1/[(3n-2)*(3n+1)]＝1/3×[1/(3n-2)－1/(3n+1)]
相加，得1/3×[1－1/(3n+1)]
所以，极限是1/3; 1年前

洼凉洼凉的心共回答了141个问题 | 采纳率: 1/3; 1年前

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1/1X4 = （1/1 - 1/4）/3
1/4X7 = (1/4 - 1/7)/3
...
1/91x94 = (1/91 - 1/94)/3
1/1X4+1/4X7+1/7X10+.+1/91x94
=(1/1-1/4 + 1/4 - 1/7 + ...+ 1/91 - 1/94)x1/3
=(1-1/94)/3
=31/94

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1/(1x4)+1/(4x7)+1/(7x10)+1/(10x13)简便运算找规律急 热zz涤灵魂1年前5: ysh2_271_d_o2974 共回答了16个问题 | 采纳率93.8%

1/(1x4)+1/(4x7)+1/(7x10)+1/(10x13)
=1/3x(1-1/4)+1/3x(1/4-1/7)+1/3x(1/7-1/10)+1/3x(1/10-1/13)
=1/3x(1-1/4+1/4-1/7+1/7-1/10+1/10-1/13)
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