数列1/(2*5) 1/(5*8) 1/(8*11)……1/[(3n_1)(3n+2)]的前几项和=___________我老师出了点方法,我就算到1/3[1/(3n-1)-1/(3n+2)]1/3(1/2-1/5)+1/3(1/5-1/8)+1/3……1/3[1/(3n-1)-1/(3n+2)]1/3[1/2-1/5+1/5-1/8+1/8-……+1/(3n-1)-1/(3n+2)]1/3[1/
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数列1/(2*5) 1/(5*8) 1/(8*11)……1/[(3n_1)(3n+2)]的前几项和=___________我老师出了点方法,我就算到1/3[1/(3n-1)-1/(3n+2)]1/3(1/2-1/5)+1/3(1/5-1/8)+1/3……1/3[1/(3n-1)-1/(3n+2)]1/3[1/2-1/5+1/5-1/8+1/8-……+1/(3n-1)-1/(3n+2)]1/3[1/
数列1/(2*5) 1/(5*8) 1/(8*11)……1/[(3n_1)(3n+2)]的前几项和=___________
我老师出了点方法,我就算到
1/3[1/(3n-1)-1/(3n+2)]
1/3(1/2-1/5)+1/3(1/5-1/8)+1/3……1/3[1/(3n-1)-1/(3n+2)]
1/3[1/2-1/5+1/5-1/8+1/8-……+1/(3n-1)-1/(3n+2)]
1/3[1/2-1/(3n+2)=___?_____
接下来怎么算啊?给个详解,就从1/3[1/2-1/(3n+2)=___?_____这个开始算给我吧!
数列1/(2*5) 1/(5*8) 1/(8*11)……1/[(3n_1)(3n+2)]的前几项和=___________我老师出了点方法,我就算到1/3[1/(3n-1)-1/(3n+2)]1/3(1/2-1/5)+1/3(1/5-1/8)+1/3……1/3[1/(3n-1)-1/(3n+2)]1/3[1/2-1/5+1/5-1/8+1/8-……+1/(3n-1)-1/(3n+2)]1/3[1/
an=1/[(3n-1)(3n+2)]=(1/3)*[1/(3n-1)-1/(3n+2)]
那么Sn=a1+a2+...+an
=(1/3)*(1/2-1/5)+(1/3)*(1/5-1/8)+...+(1/3)*[1/(3n-1)-1/(3n+2)]
=(1/3)*[1/2-1/(3n+2)]
=(1/3)*[(3n+2-2)/2(3n+2)](通分)
=n/2(3n+2)
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