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

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)