an=1/(3n-2)(3n+1),求Sn不要给我直接来个根据裂项法怎么裂的项也就是怎么变成an=1/3[1/(3n-2)-1/(3n+1)]

Question

an=1/(3n-2)(3n+1),求Sn
不要给我直接来个根据裂项法
怎么裂的项也就是怎么变成an=1/3[1/(3n-2)-1/(3n+1)]

Answer 1

1/(3n-2)-1/(3n+1)(通分）
=(3n+1)/(3n-2)(3n+1)-(3n-2)/(3n-2)(3n+1)
=(3n+1-3n+2)/(3n-2)(3n+1)
=3/(3n-2)(3n+1)
1/3*3/(3n-2)(3n+1)=1/(3n-2)(3n+1)
所以1/3*[1/(3n-2)-1/(3n+1)]=1/(3n-2)(3n+1)
Sn=1/1*4+1/4*7+1/7*10+.+1/(3n-2)(3n+1)
=1/3*(1-1/4)+1/3*(1/4-1/7)+1/3*(1/7-1/10)+.+1/3*[1/(3n-2)-1/(3n+1)]
=1/3*[1-1/4+1/4-1/7+1/7-1/10+.+1/(3n-2)-1/(3n+1)]
=1/3*[1-1/(3n+1)]
=1/3*[3n/(3n+1)]
=n/(3n+1)