数列{an}满足an+1+an=4n-3(n∈N*) (Ⅰ)若{an}是等差数列,求其通项公式; (Ⅱ)若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1.
问题描述:
数列{an}满足an+1+an=4n-3(n∈N*)
(Ⅰ)若{an}是等差数列,求其通项公式;
(Ⅱ)若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1.
答
( I)由题意得an+1+an=4n-3…①
an+2+an+1=4n+1…②.…(2分)
②-①得an+2-an=4,
∵{an}是等差数列,设公差为d,∴d=2,(4分)
∵a1+a2=1∴a1+a1+d=1,∴a1=−
.(6分)1 2
∴an=2n−
.(7分)5 2
(Ⅱ)∵a1=2,a1+a2=1,
∴a2=-1.(8分)
又∵an+2-an=4,
∴数列的奇数项与偶数项分别成等差数列,公差均为4,
∴a2n-1=4n-2,a2n=4n-5.(11分)
S2n+1=(a1+a3+…+a2n+1)+(a2+a4+…+a2n)(12分)
=(n+1)×2+
×4+n×(−1)+(n+1)n 2
×4n(n−1) 2
=4n2+n+2.(14分)