先猜想1/1*4+1/4*7+...+1/(3n-2)(3n+1)的和,并用数归法证明

1、先猜想S=1/1*4+1/4*7+...+1/(3n-2)(3n+1)=n/(3n+1)
2、证明：（1）当n=1,S=1/4
（2）假设n=k时,s=k/(3k+1)
当n=k+1时,S(k+1)=1/1*4+1/4*7+...+1/(3k-2)(3k+1)+1/(3(k+1)-2)(3(k+1)+1)=1/3*[1-1/4+1/4-1/7+1/7-1/10+...+1/(3k-2)-1/(3k+1)+1/(3(k+1)-2)-1/(3(k+1)+1)]=1/3*[1-1/(3(k+1)+1)]=3(k+1)/(3(k+1)+1) 则当n=k+1时,仍然成立.
（3）所以1/1*4+1/4*7+...+1/(3n-2)(3n+1)=n/(3n+1)

1/(3n-2)(3n+1)
=1/3*[1/(3n-2)-1/(3n+1)]
1/1*4+1/4*7+...+1/(3n-2)(3n+1)
=1/3*(1-1/4)+1/3*(1/4-17)+..........+1/3*[1/(3n-2)-1/(3n+1)]
=1/3*[1-1/4+1/4-1/7+.......+1/(3n-2)-1/(3n+1)]
=1/3*[1-1/(3n+1)]
=1/3*3n/(3n+1)
=n/(3n+1)

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