作业帮1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)+.+1/(17+18+19+20)=?1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)......+1/(17*18*19*20)=?
来源：学生作业帮助网 编辑：作业帮 时间：2024/12/01 05:49:45
1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)+.+1/(17+18+19+20)=?1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)......+1/(17*18*19*20)=?
1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)+.+1/(17+18+19+20)=?
1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)......+1/(17*18*19*20)=?
1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)+.+1/(17+18+19+20)=?1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6)......+1/(17*18*19*20)=?
1/(1*2*3*4)=1/3(1/1*2*3-1/2*3*4)
1/(2*3*4*5)=1/3(1/2*3*4-1/3*4*5)
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1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6).+1/(17*18*19*20)
=1/3(1/1*2*3-1/18*19*20)
=1/3*(3*19*20-1)/(18*19*20)
=1139/20520
分母的共性是n(n+1)(n+2)(n+3)=(n^2+3n)(n^2+3n+2).
设n^2+3n=m,题中的每一项都形如1/m(m+2)=1/2[1/m-1/(m+2)];
原式=1/2[1/1-1/(1+2)]+1/2[1/2-1/(2+2)]+…+1/2[1/17-1/(17+2)]
=1/2[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+…+1/16...
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分母的共性是n(n+1)(n+2)(n+3)=(n^2+3n)(n^2+3n+2).
设n^2+3n=m,题中的每一项都形如1/m(m+2)=1/2[1/m-1/(m+2)];
原式=1/2[1/1-1/(1+2)]+1/2[1/2-1/(2+2)]+…+1/2[1/17-1/(17+2)]
=1/2[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+…+1/16-1/18+1/17-1/19]
=1/2[1+1/2-1/18-1/19]
=119/171.
找出规律,把一项分解成两项之差,可正负抵消简化运算.
祝你不断进步.
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