数列〔an〕满足an+1+an=4n-3,当a1=2时,Sn为数列〔an〕前n项和,求S 2n+1

a(n+1) + an = 4n -3a(n+1) - 2(n+1) + 5/2 = - ( an - 2n + 5/2 )令 bn = an - 2n + 5/2b1 = a1 - 2 + 5/2 = 5/2b(n+1) = - bn∴bn = 5/2 *(-1)^(n+1)an = bn +2n - 5/2 = 5/2 [ (-1)^(n+1) - 1 ] + 2n当n为奇数时,...