Question

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．

Answer 1

a(n+1)+an=4n-3
(I)
{an}是等差数列
=> an =a1+(n-1)d

a(n+1)+an=4n-3
2a1+(2n-1)d= 4n-3
n=1, 2a1+d=1 (1)
n=2, 2a1+3d=5 (2)
(2)-(1)
d=2
from (1) =>a1=-1/2
an = -1/2 +2(n-1)
= 2n - 5/2

(II)
a1=2
a(n+1)+an=4n-3
a(n+1)-2(n+1)+5/2 = -(an -2n +5/2)

{an -2n +5/2}是等比数列, q=-1
an -2n +5/2 = (-1)^(n-1) .(a1 -2 +5/2)
=(5/2)(-1)^(n-1)
an = 2n- 5/2 +(5/2)(-1)^(n-1)

an = 2n ; if n is odd
= 2n -5 ; if n is even
S(2n+1)
= [a1+a3+...+a(2n+1) ] +[a2+a4+...+a(2n)]
=[2(2n+1)+ 2](n+1)/2 + ( 4n-5 -1)n/2
=2(n+1)(n+1) + (2n-3)n
=4n^2+n+4

Answer 2

a(n+1)+an=4n-3
an-a(n-1)=4(n-1)-3
两式相减，得到
a(n+1)-a(n-1)=4
也就是说a1,a3,a5,...a(2k-1)成等差数列，
a2,a4,a6,...a(2k)成等差数列，公差都为4，
a(2k-1)=a1+(k-1)d=2+4(k-1)=4k-2
当n=1，a2+a1=4-3=1,a2=-1
a(2k)=a2+(k-1)d=-1+4(k-1)=4k-5
an的通项公式为
an=4k-2 (n=2k-1)
4k-5 (n=2k)
希望能解决您的问题。