Question

已知数列（1/1*4），（1/4*7）…，（1/（3n-2)(3n+1)），…，计算数列和S1，S2，S3，S4，根据计算结果猜想

具体点，怎么求S1，S2...

Answer 1

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

S1=1/1*4=(1/1-1/4)/3=1/4
S2=1/1*4+1/4*7=(1/1-1/4)/3+(1/4-1/7)/3=[1-1/4+1/4-1/7]/3=[1-1/7]/3=2/7
......
Sn=1/1*4+1/4*7+......1/（3n-2)(3n+1)
=(1/1-1/4)/3+(1/4-1/7)/3+......+[1/(3n-2)-1/(3n+1)]/3
=[1-1/4+1/4-1/7+.....+1/(3n-2)-1/(3n+1)]/3
=[1-1/(3n+1)]/3
=n/(3n+1)