已知数列{an}的前几项和为Sn=3n,数列{bn}满足b1=-1,bn+1=bn+(2n-1)
问题描述:
已知数列{an}的前几项和为Sn=3n,数列{bn}满足b1=-1,bn+1=bn+(2n-1)
(n€N)……⑴求数列{an}的通项公式an⑵求数列{bn}的通项公式bn
答
(1)Sn=3n①,那么S(n-1)=3(n-1)②,①-②得an=3(2)bn+1=bn+(2n-1),那么bn+1-bn=2n-1b2-b1=2×1-1①b3-b2=2×2-1②b4-b3=2×3-1③.bn+1-bn=2n-1(n)①﹢②﹢③﹢.﹢(n)得b(n+1)-b1=2×(1+2+3+...+n)-n=n&...