(已知an=1/(3n-2)(3n+1) (n为正整数),记b1=a1,b2=a1+a2,……,bn=a1+a2……+an,则可推算出bn的表达式为

问题描述:

(已知an=1/(3n-2)(3n+1) (n为正整数),记b1=a1,b2=a1+a2,……,bn=a1+a2……+an,则可推算出bn的表达式为
n为角标

an=1/(3n-2)(3n+1)=[1/(3n-2)-1/(3n+1)]/3
b1=(1/1-1/4)/3
b2=(1/1-1/4)/3+(1/4-1/7)/3
...
bn=(1/1-1/4)/3+(1/4-1/7)/3+...+[1/(3n-2)-1/(3n+1)]/3
=[(1-1/(3n+1)]/3
=n/(3n+1)