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

( 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=−

1
2
.(6分)
an=2n−
5
2
.(7分)
(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+
(n+1)n
2
×4+n×(−1)+
n(n−1)
2
×4

=4n2+n+2.(14分)